problem solving about quadratic equation with solution

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9.6: Solve Applications of Quadratic Equations

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Learning Objectives

By the end of this section, you will be able to:

• Solve applications modeled by quadratic equations

Before you get started, take this readiness quiz.

• The sum of two consecutive odd numbers is \(−100\). Find the numbers. If you missed this problem, review Example 2.18.
• Solve: \(\frac{2}{x+1}+\frac{1}{x-1}=\frac{1}{x^{2}-1}\). If you missed this problem, review Example 7.35.
• Find the length of the hypotenuse of a right triangle with legs \(5\) inches and \(12\) inches. If you missed this problem, review Example 2.34.

Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

Methods to Solve Quadratic Equations

• Square Root Property
• Completing the Square
• Quadratic Formula

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

Use a Problem-Solving Strategy

• Read the problem. Make sure all the words and ideas are understood.
• Identify what we are looking for.
• Name what we are looking for. Choose a variable to represent that quantity.
• Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
• Solve the equation using algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is \(2\) more than the number preceding it. If we call the first one \(n\), then the next one is \(n+2\). The next one would be \(n+2+2\) or \(n+4\). This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

\(\begin{array}{cl}{}&{\text{Consecutive even integers}}\\{}& {64,66,68}\\ {n} & {1^{\text { st }} \text { even integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive even integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive even integer }}\end{array}\)

\(\begin{array}{cl}{}&{\text{Consecutive odd integers}}\\{}& {77,79,81}\\ {n} & {1^{\text { st }} \text { odd integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive odd integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive odd integer }}\end{array}\)

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

Example \(\PageIndex{1}\)

The product of two consecutive odd integers is \(195\). Find the integers.

Step 1 : Read the problem

Step 2 : Identify what we are looking for.

We are looking for two consecutive odd integers.

Step 3 : Name what we are looking for.

Let \(n=\) the first odd integer.

\(n+2=\) the next odd integer.

Step 4 : Translate into an equation. State the problem in one sentence.

“The product of two consecutive odd integers is \(195\).” The product of the first odd integer and the second odd integer is \(195\).

Translate into an equation.

\(n(n+2)=195\)

Step 5 : Solve the equation. Distribute.

\(n^{2}+2 n=195\)

Write the equation in standard form.

\(n^{2}+2 n-195=0\)

\((n+15)(n-13)=0\)

Use the Zero Product Property.

\(n+15=0 \quad n-13=0\)

Solve each equation.

\(n=-15, \quad n=13\)

There are two values of \(n\) that are solutions. This will give us two pairs of consecutive odd integers for our solution.

\(\begin{array}{cc}{\text { First odd integer } n=13} & {\text { First odd integer } n=-15} \\ {\text { next odd integer } n+2} & {\text { next odd integer } n+2} \\ {13+2} & {-15+2} \\ {15} & {-13}\end{array}\)

Step 6 : Check the answer.

Do these pairs work? Are they consecutive odd integers?

\(\begin{aligned} 13,15 & \text { yes } \\-13,-15 & \text { yes } \end{aligned}\)

Is their product \(195\)?

\(\begin{aligned} 13 \cdot 15 &=195 &\text{yes} \\-13(-15) &=195 & \text { yes } \end{aligned}\)

Step 7 : Answer the question.

Two consecutive odd integers whose product is \(195\) are \(13,15\) and \(-13,-15\).

Exercise \(\PageIndex{1}\)

The product of two consecutive odd integers is \(99\). Find the integers.

The two consecutive odd integers whose product is \(99\) are \(9, 11\), and \(−9, −11\).

Exercise \(\PageIndex{2}\)

The product of two consecutive even integers is \(168\). Find the integers.

The two consecutive even integers whose product is \(128\) are \(12, 14\) and \(−12, −14\).

We will use the formula for the area of a triangle to solve the next example.

Definition \(\PageIndex{1}\)

Area of a Triangle

For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2} b h\).

Recall that when we solve geometric applications, it is helpful to draw the figure.

Example \(\PageIndex{2}\)

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of \(120\) square feet and the architect wants the base to be \(4\) feet more than twice the height. Find the base and height of the window.

Exercise \(\PageIndex{3}\)

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of \(456\) square inches.

The height of the triangle is \(12\) inches and the base is \(76\) inches.

Exercise \(\PageIndex{4}\)

If a triangle that has an area of \(110\) square feet has a base that is two feet less than twice the height, what is the length of its base and height?

The height of the triangle is \(11\) feet and the base is \(20\) feet.

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

Definition \(\PageIndex{2}\)

Area of a Rectangle

For a rectangle with length, \(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

Example \(\PageIndex{3}\)

Mike wants to put \(150\) square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than \(3\) times the width. Find the length and width. Round to the nearest tenth of a foot.

Exercise \(\PageIndex{5}\)

The length of a \(200\) square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.

The length of the garden is approximately \(18\) feet and the width \(11\) feet.

Exercise \(\PageIndex{6}\)

A rectangular tablecloth has an area of \(80\) square feet. The width is \(5\) feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot?

The length of the tablecloth is approximately \(11.8\) feet and the width \(6.8\) feet.

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

Definition \(\PageIndex{3}\)

Pythagorean Theorem

• In any right triangle, where \(a\) and \(b\) are the lengths of the legs, and \(c\) is the length of the hypotenuse, \(a^{2}+b^{2}=c^{2}\).

Example \(\PageIndex{4}\)

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two \(10\)-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

Exercise \(\PageIndex{7}\)

The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is \(20\) feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.

The length of the flag pole’s shadow is approximately \(6.3\) feet and the height of the flag pole is \(18.9\) feet.

Exercise \(\PageIndex{8}\)

The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.

The distance between the opposite corners is approximately \(7.2\) feet.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, \(v_{0}\), propels the object up until gravity causes the object to fall back down.

Definition \(\PageIndex{4}\)

The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula

\(h=-16 t^{2}+v_{0} t\)

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

Example \(\PageIndex{5}\)

A firework is shot upwards with initial velocity \(130\) feet per second. How many seconds will it take to reach a height of \(260\) feet? Round to the nearest tenth of a second.

Exercise \(\PageIndex{9}\)

An arrow is shot from the ground into the air at an initial speed of \(108\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the arrow will be \(180\) feet from the ground. Round the nearest tenth.

The arrow will reach \(180\) feet on its way up after \(3\) seconds and again on its way down after approximately \(3.8\) seconds.

Exercise \(\PageIndex{10}\)

A man throws a ball into the air with a velocity of \(96\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the height of the ball will be \(48\) feet. Round to the nearest tenth.

The ball will reach \(48\) feet on its way up after approximately \(.6\) second and again on its way down after approximately \(5.4\) seconds.

We have solved uniform motion problems using the formula \(D=rt\) in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

The formula \(D=rt\) assumes we know \(r\) and \(t\) and use them to find \(D\). If we know \(D\) and \(r\) and need to find \(t\), we would solve the equation for \(t\) and get the formula \(t=\frac{D}{r}\).

Some uniform motion problems are also modeled by quadratic equations.

Example \(\PageIndex{6}\)

Professor Smith just returned from a conference that was \(2,000\) miles east of his home. His total time in the airplane for the round trip was \(9\) hours. If the plane was flying at a rate of \(450\) miles per hour, what was the speed of the jet stream?

This is a uniform motion situation. A diagram will help us visualize the situation.

We fill in the chart to organize the information.

We are looking for the speed of the jet stream. Let \(r=\) the speed of the jet stream.

When the plane flies with the wind, the wind increases its speed and so the rate is \(450 + r\).

When the plane flies against the wind, the wind decreases its speed and the rate is \(450 − r\).

The speed of the jet stream was \(50\) mph.

Exercise \(\PageIndex{11}\)

MaryAnne just returned from a visit with her grandchildren back east. The trip was \(2400\) miles from her home and her total time in the airplane for the round trip was \(10\) hours. If the plane was flying at a rate of \(500\) miles per hour, what was the speed of the jet stream?

The speed of the jet stream was \(100\) mph.

Exercise \(\PageIndex{12}\)

Gerry just returned from a cross country trip. The trip was \(3000\) miles from his home and his total time in the airplane for the round trip was \(11\) hours. If the plane was flying at a rate of \(550\) miles per hour, what was the speed of the jet stream?

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We’ll use a similar scenario now.

Example \(\PageIndex{7}\)

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(12\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(8\) hours. How long does it take for each press to print the job alone?

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

Exercise \(\PageIndex{13}\)

The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(6\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(4\) hours. How long does it take for each press to print the job alone?

Press #1 would take \(12\) hours, and Press #2 would take \(6\) hours to do the job alone.

Exercise \(\PageIndex{14}\)

Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes \(3\) hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in \(2\) hours. How long does it take for each hose to fill the hot tub?

The red hose take \(6\) hours and the green hose take \(3\) hours alone.

Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.

• Word Problems Involving Quadratic Equations
• Quadratic Equation Word Problems
• Applying the Quadratic Formula

Key Concepts

• Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
• Solve the equation using good algebra techniques.
• For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2}bh\).

• For a rectangle with length,\(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

• The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula \(h=-16 t^{2}+v_{0} t\).

• Math Article

Quadratic Equation Questions

Quadratic equation questions are provided here for Class 10 students. A quadratic equation is a second-degree polynomial which is represented as ax 2 + bx + c = 0, where a is not equal to 0. Here, a, b and c are constants, also called coefficients and x is an unknown variable. Also, learn Quadratic Formula here.

Solving the problems based on quadratics will help students to understand the concept very well and also to score good marks in this section. All the questions are solved here step by step with a detailed explanation. In this article, we will give the definition and important formula for solving problems based on quadratic equations. The questions given here is in reference to the CBSE syllabus and NCERT curriculum. 

Definition of Quadratic Equation

Usually, the quadratic equation is represented in the form of ax 2 +bx+c=0, where x is the variable and a,b,c are the real numbers & a ≠ 0. Here, a and b are the coefficients of x 2 and x, respectively. So, basically, a quadratic equation is a polynomial whose highest degree is 2. Let us see some examples:

• 3x 2 +x+1, where a=3, b=1, c=1
• 9x 2 -11x+5, where a=9, b=-11, c=5

Roots of Quadratic Equations:

If we solve any quadratic equation, then the value we obtain are called the roots of the equation. Since the degree of the quadratic equation is two, therefore we get here two solutions and hence two roots.

There are different methods to find the roots of quadratic equation, such as:

• Factorisation
• Completing the square
• Using quadratic formula

Learn: Factorization of Quadratic equations

Quadratic Equation Formula:

The quadratic formula to find the roots of the quadratic equation is given by:

\(\begin{array}{l}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\end{array} \)

Where b 2 -4ac is called the discriminant of the equation.

Based on the discriminant value, there are three possible conditions, which defines the nature of roots as follows:

• two distinct real roots, if b 2 – 4ac > 0
• two equal real roots, if b 2 – 4ac = 0
• no real roots, if b 2 – 4ac < 0

Also, learn quadratic equations for class 10 here.

Quadratic Equations Problems and Solutions

1. Rahul and Rohan have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they both have now is 124. How to find out how many marbles they had to start with.

Solution: Say, the number of marbles Rahul had be x.

Then the number of marbles Rohan had = 45 – x.

The number of marbles left with Rahul after losing 5 marbles = x – 5

The number of marbles left with Rohan after losing 5 marbles = 45 – x – 5 = 40 – x

The product of number of marbles = 124

(x – 5) (40 – x) = 124

40x – x 2 – 200 + 5x = 124

– x 2 + 45x – 200 = 124

x 2 – 45x + 324 = 0

This represents the quadratic equation. Hence by solving the given equation for x, we get;

x = 36 and x = 9

So, the number of marbles Rahul had is 36 and Rohan had is 9 or vice versa.

2. Check if x(x + 1) + 8 = (x + 2) (x – 2) is in the form of quadratic equation.

Solution: Given,

x(x + 1) + 8 = (x + 2) (x – 2)

Cancel x 2 both the sides.

Since, this expression is not in the form of ax 2 +bx+c, hence it is not a quadratic equation.

3. Find the roots of the equation 2x 2 – 5x + 3 = 0 using factorisation.

2x 2 – 5x + 3 = 0

2x 2 – 2x-3x+3 = 0

2x(x-1)-3(x-1) = 0

(2x-3) (x-1) = 0

2x-3 = 0; x = 3/2

(x-1) = 0; x=1

Therefore, 3/2 and 1 are the roots of the given equation.

4. Solve the quadratic equation 2x 2 + x – 300 = 0 using factorisation.

Solution: 2x 2 + x – 300 = 0

2x 2 – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

(x – 12)(2x + 25) = 0

x-12=0; x=12

(2x+25) = 0; x=-25/2 = -12.5

Therefore, 12 and -12.5 are two roots of the given equation.

Also, read   Factorisation .

5. Solve the equation x 2 +4x-5=0.

x 2 + 4x – 5 = 0

x 2 -1x+5x-5 = 0

x(x-1)+5(x-1) =0

(x-1)(x+5) =0

Hence, (x-1) =0, and (x+5) =0

similarly, x+5 = 0

x=-5 & x=1

6. Solve the quadratic equation 2x 2 + x – 528 = 0, using quadratic formula.

Solution: If we compare it with standard equation, ax 2 +bx+c = 0

a=2, b=1 and c=-528

Hence, by using the quadratic formula:

Now putting the values of a,b and c.

x=64/4 or x=-66/4

x=16 or x=-33/2

7. Find the roots of x 2 + 4x + 5 = 0, if any exist, using quadratic formula.

Solution: To check whether there are real roots available for the quadratic equation, we need the find the discriminant value.

D = b 2 -4ac = 4 2 – 4(1)(5) = 16-20 = -4

Since the square root of -4 will not give a real number. Hence there is no real roots for the given equation.

8. Find the discriminant of the equation: 3x 2 -2x+⅓ = 0.

Solution: Here, a = 3, b=-2 and c=⅓

Hence, discriminant, D = b 2 – 4ac

D = (-2) 2 -4(3)(⅓)

Video Lesson

Quadratic equation worksheet.

Practice Questions

Solve these quadratic equations and find the roots. 

• x 2 -5x-14=0         [Answer: x=-2 & x=7]
• X 2  = 11x -28       [Answer: x=4 & x = 7]
• 6x 2 – x = 5            [Answer: x=-⅚ & x = 1]
• 12x 2 = 25x          [Answer: x=0 & x=25/12]

Frequently Asked Questions on Quadratic Equations

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10.3 Solve Quadratic Equations Using the Quadratic Formula

Learning objectives.

By the end of this section, you will be able to:

• Solve quadratic equations using the quadratic formula
• Use the discriminant to predict the number of solutions of a quadratic equation
• Identify the most appropriate method to use to solve a quadratic equation

Be Prepared 10.7

Before you get started, take this readiness quiz.

Simplify: −20 − 5 10 −20 − 5 10 . If you missed this problem, review Example 1.74 .

Be Prepared 10.8

Simplify: 4 + 121 4 + 121 . If you missed this problem, review Example 9.29 .

Be Prepared 10.9

Simplify: 128 128 . If you missed this problem, review Example 9.12 .

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes.’ In this section, we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’ so that we would do the algebraic steps only once and then use the new formula to find the value of the specific variable. Now, we will go through the steps of completing the square in general to solve a quadratic equation for x . It may be helpful to look at one of the examples at the end of the last section where we solved an equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 as you read through the algebraic steps below, so you see them with numbers as well as ‘in general.’

This last equation is the Quadratic Formula.

• Quadratic Formula

The solutions to a quadratic equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 , a ≠ 0 a ≠ 0 are given by the formula:

To use the Quadratic Formula, we substitute the values of a , b , and c a , b , and c into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation.

Example 10.28

How to solve a quadratic equation using the quadratic formula.

Solve 2 x 2 + 9 x − 5 = 0 2 x 2 + 9 x − 5 = 0 by using the Quadratic Formula.

Try It 10.55

Solve 3 y 2 − 5 y + 2 = 0 3 y 2 − 5 y + 2 = 0 by using the Quadratic Formula.

Try It 10.56

Solve 4 z 2 + 2 z − 6 = 0 4 z 2 + 2 z − 6 = 0 by using the Quadratic Formula.

Solve a quadratic equation using the Quadratic Formula.

• Step 1. Write the Quadratic Formula in standard form. Identify the a a , b b , and c c values.
• Step 2. Write the Quadratic Formula. Then substitute in the values of a a , b b , and c . c .
• Step 3. Simplify.
• Step 4. Check the solutions.

If you say the formula as you write it in each problem, you’ll have it memorized in no time. And remember, the Quadratic Formula is an equation. Be sure you start with ‘ x = x = ’.

Example 10.29

Solve x 2 − 6 x + 5 = 0 x 2 − 6 x + 5 = 0 by using the Quadratic Formula.

Try It 10.57

Solve a 2 − 2 a − 15 = 0 a 2 − 2 a − 15 = 0 by using the Quadratic Formula.

Try It 10.58

Solve b 2 + 10 b + 24 = 0 b 2 + 10 b + 24 = 0 by using the Quadratic Formula.

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form.

Example 10.30

Solve 4 y 2 − 5 y − 3 = 0 4 y 2 − 5 y − 3 = 0 by using the Quadratic Formula.

We can use the Quadratic Formula to solve for the variable in a quadratic equation, whether or not it is named ‘ x ’.

Try It 10.59

Solve 2 p 2 + 8 p + 5 = 0 2 p 2 + 8 p + 5 = 0 by using the Quadratic Formula.

Try It 10.60

Solve 5 q 2 − 11 q + 3 = 0 5 q 2 − 11 q + 3 = 0 by using the Quadratic Formula.

Example 10.31

Solve 2 x 2 + 10 x + 11 = 0 2 x 2 + 10 x + 11 = 0 by using the Quadratic Formula.

Try It 10.61

Solve 3 m 2 + 12 m + 7 = 0 3 m 2 + 12 m + 7 = 0 by using the Quadratic Formula.

Try It 10.62

Solve 5 n 2 + 4 n − 4 = 0 5 n 2 + 4 n − 4 = 0 by using the Quadratic Formula.

We cannot take the square root of a negative number. So, when we substitute a a , b b , and c c into the Quadratic Formula, if the quantity inside the radical is negative, the quadratic equation has no real solution. We will see this in the next example.

Example 10.32

Solve 3 p 2 + 2 p + 9 = 0 3 p 2 + 2 p + 9 = 0 by using the Quadratic Formula.

Try It 10.63

Solve 4 a 2 − 3 a + 8 = 0 4 a 2 − 3 a + 8 = 0 by using the Quadratic Formula.

Try It 10.64

Solve 5 b 2 + 2 b + 4 = 0 5 b 2 + 2 b + 4 = 0 by using the Quadratic Formula.

The quadratic equations we have solved so far in this section were all written in standard form, a x 2 + b x + c = 0 a x 2 + b x + c = 0 . Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Example 10.33

Solve x ( x + 6 ) + 4 = 0 x ( x + 6 ) + 4 = 0 by using the Quadratic Formula.

Try It 10.65

Solve x ( x + 2 ) − 5 = 0 x ( x + 2 ) − 5 = 0 by using the Quadratic Formula.

Try It 10.66

Solve y ( 3 y − 1 ) − 2 = 0 y ( 3 y − 1 ) − 2 = 0 by using the Quadratic Formula.

When we solved linear equations, if an equation had too many fractions we ‘cleared the fractions’ by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions—to solve. We can use the same strategy with quadratic equations.

Example 10.34

Solve 1 2 u 2 + 2 3 u = 1 3 1 2 u 2 + 2 3 u = 1 3 by using the Quadratic Formula.

Try It 10.67

Solve 1 4 c 2 − 1 3 c = 1 12 1 4 c 2 − 1 3 c = 1 12 by using the Quadratic Formula.

Try It 10.68

Solve 1 9 d 2 − 1 2 d = − 1 2 1 9 d 2 − 1 2 d = − 1 2 by using the Quadratic Formula.

Think about the equation ( x − 3 ) 2 = 0 ( x − 3 ) 2 = 0 . We know from the Zero Products Principle that this equation has only one solution: x = 3 x = 3 .

We will see in the next example how using the Quadratic Formula to solve an equation with a perfect square also gives just one solution.

Example 10.35

Solve 4 x 2 − 20 x = −25 4 x 2 − 20 x = −25 by using the Quadratic Formula.

Did you recognize that 4 x 2 − 20 x + 25 4 x 2 − 20 x + 25 is a perfect square?

Try It 10.69

Solve r 2 + 10 r + 25 = 0 r 2 + 10 r + 25 = 0 by using the Quadratic Formula.

Try It 10.70

Solve 25 t 2 − 40 t = −16 25 t 2 − 40 t = −16 by using the Quadratic Formula.

Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two solutions, sometimes one solution, sometimes no real solutions. Is there a way to predict the number of solutions to a quadratic equation without actually solving the equation?

Yes, the quantity inside the radical of the Quadratic Formula makes it easy for us to determine the number of solutions. This quantity is called the discriminant .

Discriminant

In the Quadratic Formula x = − b ± b 2 − 4 a c 2 a x = − b ± b 2 − 4 a c 2 a , the quantity b 2 − 4 a c b 2 − 4 a c is called the discriminant .

Let’s look at the discriminant of the equations in Example 10.28 , Example 10.32 , and Example 10.35 , and the number of solutions to those quadratic equations.

When the discriminant is positive ( x = − b ± + 2 a ) ( x = − b ± + 2 a ) the quadratic equation has two solutions .

When the discriminant is zero ( x = − b ± 0 2 a ) ( x = − b ± 0 2 a ) the quadratic equation has one solution .

When the discriminant is negative ( x = − b ± − 2 a ) ( x = − b ± − 2 a ) the quadratic equation has no real solutions .

Use the discriminant, b 2 − 4 a c b 2 − 4 a c , to determine the number of solutions of a Quadratic Equation.

For a quadratic equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 , a ≠ 0 a ≠ 0 ,

• if b 2 − 4 a c > 0 b 2 − 4 a c > 0 , the equation has two solutions.
• if b 2 − 4 a c = 0 b 2 − 4 a c = 0 , the equation has one solution.
• if b 2 − 4 a c < 0 b 2 − 4 a c < 0 , the equation has no real solutions.

Example 10.36

Determine the number of solutions to each quadratic equation:

ⓐ 2 v 2 − 3 v + 6 = 0 2 v 2 − 3 v + 6 = 0 ⓑ 3 x 2 + 7 x − 9 = 0 3 x 2 + 7 x − 9 = 0 ⓒ 5 n 2 + n + 4 = 0 5 n 2 + n + 4 = 0 ⓓ 9 y 2 − 6 y + 1 = 0 9 y 2 − 6 y + 1 = 0

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

Try It 10.71

ⓐ 8 m 2 − 3 m + 6 = 0 8 m 2 − 3 m + 6 = 0 ⓑ 5 z 2 + 6 z − 2 = 0 5 z 2 + 6 z − 2 = 0 ⓒ 9 w 2 + 24 w + 16 = 0 9 w 2 + 24 w + 16 = 0 ⓓ 9 u 2 − 2 u + 4 = 0 9 u 2 − 2 u + 4 = 0

Try It 10.72

ⓐ b 2 + 7 b − 13 = 0 b 2 + 7 b − 13 = 0 ⓑ 5 a 2 − 6 a + 10 = 0 5 a 2 − 6 a + 10 = 0 ⓒ 4 r 2 − 20 r + 25 = 0 4 r 2 − 20 r + 25 = 0 ⓓ 7 t 2 − 11 t + 3 = 0 7 t 2 − 11 t + 3 = 0

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We have used four methods to solve quadratic equations:

• Square Root Property
• Completing the Square

You can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method to use.

Identify the most appropriate method to solve a Quadratic Equation.

• Step 1. Try Factoring first. If the quadratic factors easily, this method is very quick.
• Step 2. Try the Square Root Property next. If the equation fits the form a x 2 = k a x 2 = k or a ( x − h ) 2 = k a ( x − h ) 2 = k , it can easily be solved by using the Square Root Property.
• Step 3. Use the Quadratic Formula . Any quadratic equation can be solved by using the Quadratic Formula.

What about the method of completing the square? Most people find that method cumbersome and prefer not to use it. We needed to include it in this chapter because we completed the square in general to derive the Quadratic Formula. You will also use the process of completing the square in other areas of algebra.

Example 10.37

Identify the most appropriate method to use to solve each quadratic equation:

ⓐ 5 z 2 = 17 5 z 2 = 17 ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0 ⓒ 8 u 2 + 6 u = 11 8 u 2 + 6 u = 11

ⓐ 5 z 2 = 17 5 z 2 = 17

Since the equation is in the a x 2 = k a x 2 = k , the most appropriate method is to use the Square Root Property.

ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0

We recognize that the left side of the equation is a perfect square trinomial, and so Factoring will be the most appropriate method.

ⓒ 8 u 2 + 6 u = 11 8 u 2 + 6 u = 11

Put the equation in standard form. 8 u 2 + 6 u − 11 = 0 8 u 2 + 6 u − 11 = 0

While our first thought may be to try Factoring, thinking about all the possibilities for trial and error leads us to choose the Quadratic Formula as the most appropriate method

Try It 10.73

ⓐ x 2 + 6 x + 8 = 0 x 2 + 6 x + 8 = 0 ⓑ ( n − 3 ) 2 = 16 ( n − 3 ) 2 = 16 ⓒ 5 p 2 − 6 p = 9 5 p 2 − 6 p = 9

Try It 10.74

ⓐ 8 a 2 + 3 a − 9 = 0 8 a 2 + 3 a − 9 = 0 ⓑ 4 b 2 + 4 b + 1 = 0 4 b 2 + 4 b + 1 = 0 ⓒ 5 c 2 = 125 5 c 2 = 125

Access these online resources for additional instruction and practice with using the Quadratic Formula:

• Solving Quadratic Equations: Solving with the Quadratic Formula
• How to solve a quadratic equation in standard form using the Quadratic Formula (example)
• Solving Quadratic Equations using the Quadratic Formula—Example 3
• Solve Quadratic Equations using Quadratic Formula

Section 10.3 Exercises

Practice makes perfect.

Solve Quadratic Equations Using the Quadratic Formula

In the following exercises, solve by using the Quadratic Formula.

4 m 2 + m − 3 = 0 4 m 2 + m − 3 = 0

4 n 2 − 9 n + 5 = 0 4 n 2 − 9 n + 5 = 0

2 p 2 − 7 p + 3 = 0 2 p 2 − 7 p + 3 = 0

3 q 2 + 8 q − 3 = 0 3 q 2 + 8 q − 3 = 0

p 2 + 7 p + 12 = 0 p 2 + 7 p + 12 = 0

q 2 + 3 q − 18 = 0 q 2 + 3 q − 18 = 0

r 2 − 8 r − 33 = 0 r 2 − 8 r − 33 = 0

t 2 + 13 t + 40 = 0 t 2 + 13 t + 40 = 0

3 u 2 + 7 u − 2 = 0 3 u 2 + 7 u − 2 = 0

6 z 2 − 9 z + 1 = 0 6 z 2 − 9 z + 1 = 0

2 a 2 − 6 a + 3 = 0 2 a 2 − 6 a + 3 = 0

5 b 2 + 2 b − 4 = 0 5 b 2 + 2 b − 4 = 0

2 x 2 + 3 x + 9 = 0 2 x 2 + 3 x + 9 = 0

6 y 2 − 5 y + 2 = 0 6 y 2 − 5 y + 2 = 0

v ( v + 5 ) − 10 = 0 v ( v + 5 ) − 10 = 0

3 w ( w − 2 ) − 8 = 0 3 w ( w − 2 ) − 8 = 0

1 3 m 2 + 1 12 m = 1 4 1 3 m 2 + 1 12 m = 1 4

1 3 n 2 + n = − 1 2 1 3 n 2 + n = − 1 2

16 c 2 + 24 c + 9 = 0 16 c 2 + 24 c + 9 = 0

25 d 2 − 60 d + 36 = 0 25 d 2 − 60 d + 36 = 0

5 m 2 + 2 m − 7 = 0 5 m 2 + 2 m − 7 = 0

8 n 2 − 3 n + 3 = 0 8 n 2 − 3 n + 3 = 0

p 2 − 6 p − 27 = 0 p 2 − 6 p − 27 = 0

25 q 2 + 30 q + 9 = 0 25 q 2 + 30 q + 9 = 0

4 r 2 + 3 r − 5 = 0 4 r 2 + 3 r − 5 = 0

3 t ( t − 2 ) = 2 3 t ( t − 2 ) = 2

2 a 2 + 12 a + 5 = 0 2 a 2 + 12 a + 5 = 0

4 d 2 − 7 d + 2 = 0 4 d 2 − 7 d + 2 = 0

3 4 b 2 + 1 2 b = 3 8 3 4 b 2 + 1 2 b = 3 8

1 9 c 2 + 2 3 c = 3 1 9 c 2 + 2 3 c = 3

2 x 2 + 12 x − 3 = 0 2 x 2 + 12 x − 3 = 0

16 y 2 + 8 y + 1 = 0 16 y 2 + 8 y + 1 = 0

In the following exercises, determine the number of solutions to each quadratic equation.

• ⓐ 4 x 2 − 5 x + 16 = 0 4 x 2 − 5 x + 16 = 0
• ⓑ 36 y 2 + 36 y + 9 = 0 36 y 2 + 36 y + 9 = 0
• ⓒ 6 m 2 + 3 m − 5 = 0 6 m 2 + 3 m − 5 = 0
• ⓓ 18 n 2 − 7 n + 3 = 0 18 n 2 − 7 n + 3 = 0
• ⓐ 9 v 2 − 15 v + 25 = 0 9 v 2 − 15 v + 25 = 0
• ⓑ 100 w 2 + 60 w + 9 = 0 100 w 2 + 60 w + 9 = 0
• ⓒ 5 c 2 + 7 c − 10 = 0 5 c 2 + 7 c − 10 = 0
• ⓓ 15 d 2 − 4 d + 8 = 0 15 d 2 − 4 d + 8 = 0
• ⓐ r 2 + 12 r + 36 = 0 r 2 + 12 r + 36 = 0
• ⓑ 8 t 2 − 11 t + 5 = 0 8 t 2 − 11 t + 5 = 0
• ⓒ 4 u 2 − 12 u + 9 = 0 4 u 2 − 12 u + 9 = 0
• ⓓ 3 v 2 − 5 v − 1 = 0 3 v 2 − 5 v − 1 = 0
• ⓐ 25 p 2 + 10 p + 1 = 0 25 p 2 + 10 p + 1 = 0
• ⓑ 7 q 2 − 3 q − 6 = 0 7 q 2 − 3 q − 6 = 0
• ⓒ 7 y 2 + 2 y + 8 = 0 7 y 2 + 2 y + 8 = 0
• ⓓ 25 z 2 − 60 z + 36 = 0 25 z 2 − 60 z + 36 = 0

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.

ⓐ x 2 − 5 x − 24 = 0 x 2 − 5 x − 24 = 0 ⓑ ( y + 5 ) 2 = 12 ( y + 5 ) 2 = 12 ⓒ 14 m 2 + 3 m = 11 14 m 2 + 3 m = 11

ⓐ ( 8 v + 3 ) 2 = 81 ( 8 v + 3 ) 2 = 81 ⓑ w 2 − 9 w − 22 = 0 w 2 − 9 w − 22 = 0 ⓒ 4 n 2 − 10 = 6 4 n 2 − 10 = 6

ⓐ 6 a 2 + 14 = 20 6 a 2 + 14 = 20 ⓑ ( x − 1 4 ) 2 = 5 16 ( x − 1 4 ) 2 = 5 16 ⓒ y 2 − 2 y = 8 y 2 − 2 y = 8

ⓐ 8 b 2 + 15 b = 4 8 b 2 + 15 b = 4 ⓑ 5 9 v 2 − 2 3 v = 1 5 9 v 2 − 2 3 v = 1 ⓒ ( w + 4 3 ) 2 = 2 9 ( w + 4 3 ) 2 = 2 9

Everyday Math

A flare is fired straight up from a ship at sea. Solve the equation 16 ( t 2 − 13 t + 40 ) = 0 16 ( t 2 − 13 t + 40 ) = 0 for t t , the number of seconds it will take for the flare to be at an altitude of 640 feet.

An architect is designing a hotel lobby. She wants to have a triangular window looking out to an atrium, with the width of the window 6 feet more than the height. Due to energy restrictions, the area of the window must be 140 square feet. Solve the equation 1 2 h 2 + 3 h = 140 1 2 h 2 + 3 h = 140 for h h , the height of the window.

Writing Exercises

Solve the equation x 2 + 10 x = 200 x 2 + 10 x = 200 ⓐ by completing the square ⓑ using the Quadratic Formula ⓒ Which method do you prefer? Why?

Solve the equation 12 y 2 + 23 y = 24 12 y 2 + 23 y = 24 ⓐ by completing the square ⓑ using the Quadratic Formula ⓒ Which method do you prefer? Why?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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Solving quadratic equation

Here you will learn about solving quadratic equations and how to do it using a graph, factoring the equation or using the quadratic formula.

Students will first learn about solving quadratic equations as part of algebra in high school.

What is solving quadratic equations?

Solving quadratic equations is finding the roots or the x- intercepts of the parabola formed by a quadratic equation.

For example,

Solve 2x^2+7x-4 by using its graph below.

The parabola passes through the x- axis as (-4,0) and \left(\cfrac{1}{2}, 0\right).

The solutions are x=-4 and x=\cfrac{1}{2}.

Solve 2x^2+7x-4=0 by factoring.

Find where each value being multiplied will be equal to 0.

2x-1=0, so x=\cfrac{1}{2}

x+4=0 so x=-4

Solve 2x^2+7x-4=0 using the quadratic formula.

The quadratic formula uses the values from the constant terms a, b and c when quadratic equations are in the general form ax^2+b x+c=0.

\begin{aligned} & x=\cfrac{-b+\sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-7+\sqrt{7^2-4 \times 2 \times(-4)}}{2 \times 2} \\\\ & x=\cfrac{-7+\sqrt{49-(-32)}}{4} \\\\ & x=\cfrac{-7+9}{4} \\\\ & x=\cfrac{2}{4} \\\\ & x=\cfrac{1}{2} \end{aligned} \quad \begin{aligned} & x=\cfrac{-b-\sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-7-\sqrt{7^2-4 \times 2 \times(-4)}}{2 \times 2} \\\\ & x=\cfrac{-7-\sqrt{49-(-32)}}{4} \\\\ & x=\cfrac{-7-9}{4} \\\\ & x=\cfrac{-16}{4} \\\\ & x=-4 \end{aligned}

Common Core State Standards

How does this relate to high school math?

• Algebra – Reasoning with Equations and Inequalities (HSA.REI.B.4b) Solve quadratic equations in one variable. Solve quadratic equations by inspection ( e.g., for x^2=49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a \pm bi for real numbers a and b.

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How to solve a quadratic equation by graphing

In order to solve quadratic equations by graphing:

Graph the quadratic equation.

Identify the \textbf{x}- intercepts.

Solving quadratic equation examples

Example 1: solve a quadratic equation by looking at its graph.

Solve the following equation by using a graph: x^2-x-6.

2 Identify the \textbf{x}- intercepts.

The parabola passes through the x- axis as (-2,0) and (3,0)

The solutions are x=-2 and x=3.

Example 2: solve a quadratic equation by looking at its graph

Solve the following equation by using a graph: 2x^2-3x+1.125.

The parabola passes through the x- axis as (0.75,0). Since it crosses at the vertex of the parabola, there is only 1 solution.

The solution is x=0.75.

See also : Quadratic graphs

How to solve a quadratic equation by factoring

In order to solve quadratic equations by factoring:

Make sure that the equation is equal to \bf{0}.

• Fully factor the quadratic equation .
• Set each expression equal to \bf{0}.

Solve for \textbf{x}.

Example 3: solve a quadratic equation by factoring with coefficient a=1

Solve x^{2}-8x+15=0 by factoring.

Fully factor the quadratic equation

(x-3)(x-5)=0

Set each expression equal to \textbf{0}.

x-3=0 \quad x-5=0

The opposite of -3 is +3, so add 3 to both sides of the equation.

The opposite of -5 is +5, so add 5 to both sides of the equation.

You can see the real roots of the quadratic equation are where the quadratic graph crosses the x- axis.

Example 4: solve a quadratic equation by factoring with coefficient a>1

Solve 2x^{2}+5x+3=0 by factoring.

(2x+3)(x+1)=0

2 x+3=0 \quad x+1=0

The opposite of +3 is -3, so subtract 3 to both sides of the equation.

The opposite of \times 2 is \div 2, so divide by 2 on both sides of the equation.

The opposite of +1 is -1, so -1 to both sides of the equation.

How to solve quadratic equations with the quadratic formula

In order to solve quadratic equations with the quadratic formula:

Identify the value of \bf{\textbf{a}, \textbf{b}} and \textbf{c} in a quadratic equation.

Substitute these values into the quadratic formula.

Solve the equation with a \bf{+}, and then with a \bf{-}.

Example 5: equation of the form ax²+b x+c=0 where a = 1

Solve x^2-2 x-15=0 using the quadratic formula.

\begin{aligned} & x=\cfrac{-(-2)+\sqrt{(-2)^2-4(1)(-15)}}{2(1)} \\\\ & x=\cfrac{2+\sqrt{4-(-60)}}{2} \\\\ & x=\cfrac{2+8}{2} \\\\ & x=5 \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-2)-\sqrt{(-2)^2-4(1)(-15)}}{2(1)} \\\\ & x=\cfrac{2-\sqrt{4-(-60)}}{2} \\\\ & x=\cfrac{2-8}{2} \\\\ & x=-3 \end{aligned}

You can see the roots of the quadratic equation are where the quadratic graph crosses the x- axis.

You can also check that the solutions are correct by substituting them into the original equation.

Example 6: solutions of the quadratic equation with a<1

Solve -6 x^2+2 x-x=-2 using the quadratic formula.

First, simplify the equation so that it is in the form ax^2+b x+c=0, which leaves a quadratic expression on the left side of the equation and 0 on the right side of the equation.

-6 x^2+x=-2 \hspace{0.5cm} *Combine the like terms (2x-x)

-6 x^2+x=-2 \hspace{0.5cm} *Add 2 to both sides so that the equation equals 0

\hspace{0.8cm} +2 \hspace{0.4cm}+2

-6 x^2+x+2=0

\begin{aligned} & x=\cfrac{-(1)+\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1+\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1+7}{-12} \\\\ & x=-\cfrac{1}{2} \end{aligned} \quad \begin{aligned} & x=\cfrac{-(1)-\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1-\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1-7}{-12} \\\\ & x=\cfrac{2}{3} \end{aligned}

Teaching tips for solving quadratic equations

• Let students use a graphing calculator to explore how changing the values for a, b and c in a quadratic equation changes the parabola.
• When students start using the quadratic formula, still encourage them to explain the meaning of their solutions. This way they do not forget what exactly they are solving for. One way to do this is to incorporate many real world examples of the quadratic equation that require students to explain the roots within the context of the problem.

Easy mistakes to make

• Thinking the order of the factored equation matters When you multiply two values the order doesn’t matter. For example, 2\times 3=3\times 2 It is the same with quadratics. (x-6)(x+4) means (x-6)\times (x+4) So, (x-6)(x+4)=0 is the same as (x+4)(x-6)=0.
• Forgetting to solve after factoring Don’t forget to set the factored expression equal to zero and solve it. Always check you have answered the question.

• Thinking an equation that cannot be factored has no solutions If a quadratic equation cannot be factored, you can still solve it by using the quadratic formula. To work out the number of real solutions a quadratic equation has you can use the discriminant.
• Not converting to the form \bf{\textbf{ax}^{2}+\textbf{b x}+\textbf{c}=0} before trying to factor While it is possible to factor a quadratic equation without it in the standard form, this can be challenging. It is recommended to use the general form shown.

Related quadratic equation lessons

• Quadratic formula
• Solving quadratic equations by graphing
• Factoring quadratic equations

Solving quadratic equations questions (by factoring)

1) Solve -x^2+2x by using its graph below.

The parabola passes through the x- axis as (0,0) and (2,0).

The solutions are x = 0 and x = 2.

2) Solve x^2+10 x+9 by using its graph below.

The parabola passes through the x- axis as (-1,0) and (-9,0).

The solutions are x = -1 and x = -9.

3) Solve {x}^2+5x+6=0 by factoring.

{x}^2+5x+6=0 can be factored as (x+2)(x+3).

Setting each factor equal to zero and solving leads to the solutions.

\begin{aligned} & x+2=0 \\\\ & x=-2 \end{aligned} \quad \begin{aligned} & x+3=0 \\\\ & x=-3 \end{aligned}

4) Solve {x}^2-x-20=0 by factoring.

x^2-x-20 can be factored as (x-5)(x+4).

\begin{aligned} & x-5=0 \\\\ & x=5 \end{aligned} \quad \begin{aligned} & x+4=0 \\\\ & x=-4 \end{aligned}

5) Solve 2{x}^2+3x-9=0 using the quadratic equation.

x=-\cfrac{3}{2}=-1.54 and x= -3

x=\cfrac{3}{2}=1.5 and x=3

x=-\cfrac{3}{2}=-1.5 and x=3

x=\cfrac{3}{2}=1.5 and x=-3

Identify a, b and c.

Solve the equation with a +, and then with a -.

\begin{aligned} & x=\cfrac{-3+\sqrt{(3)^2-4(2)(-9)}}{2(2)} \\\\ & x=\cfrac{-3+\sqrt{9-(-72)}}{4} \\\\ & x=\cfrac{-3+9}{4} \\\\ & x=\cfrac{3}{2}=1.5 \end{aligned} \quad \begin{aligned} & x=\cfrac{-3-\sqrt{(3)^2-4(2)(-9)}}{2(2)} \\\\ & x=\cfrac{-3-\sqrt{9-(-72)}}{4} \\\\ & x=\cfrac{-3-9}{4} \\\\ & x=-3 \end{aligned}

*Divide the numerator by the denominator to convert \cfrac{3}{2} to decimal form.

6) Solve 3{x}^2-9x+6=0 using the quadratic equation.

\begin{aligned} & x=\cfrac{-(-9)+\sqrt{(-9)^2-4(3)(6)}}{2(3)} \\\\ & x=\cfrac{9+\sqrt{81-72}}{6} \\\\ & x=\cfrac{9+3}{6} \\\\ & x=2 \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-9)-\sqrt{(-9)^2-4(3)(6)}}{2(3)} \\\\ & x=\cfrac{9-\sqrt{81-72}}{6} \\\\ & x=\cfrac{9-3}{6} \\\\ & x=1 \end{aligned}

Solving quadratic equation FAQs

It is a polynomial equation whose highest variable is to the second-degree ( exponent of 2).

The standard form is ax^2+b x+c=0 or f(x)=a x^2+b x+c. The coefficients a, b and c can be whole numbers, integers, fractions, decimals or any other real number where a ≠ 0.

It is the value below the radical in the quadratic formula. If the discriminant b^2-4ac is positive, there are two solutions. If it is 0, there is 1 solution. If it is negative, there are no real solutions.

The next lessons are

• Angles in parallel lines

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Solving Quadratic Equations

Before going to learn about solving quadratic equations, let us recall a few facts about quadratic equations. The word "quadratic" is originated from the word "quad" and its meaning is "square". It means the quadratic equation has a variable raised to 2 as the greatest power term. The standard form of a quadratic equation is given by the equation ax 2 + bx + c = 0, where a ≠ 0. We know that any value(s) of x that satisfies the equation is known as a solution (or) root of the equation and the process of finding the values of x which satisfy the equation ax 2 + bx + c = 0 is known as solving quadratic equations.

There are different methods used to solve quadratic equations. But the most popular method is solving quadratic equations by factoring. Let us learn all the methods in detail here along with a few solved examples.

How to Solve Quadratic Equations?

Solving quadratic equations means finding a value (or) values of variable which satisfy the equation. The value(s) that satisfy the quadratic equation is known as its roots (or) solutions (or) zeros. Since the degree of the quadratic equation is 2, it can have a maximum of 2 roots. For example, one can easily see that x = 1 and x = 2 satisfy the quadratic equation x 2 - 3x + 2 = 0 (you can substitute each of the values in this equation and verify). Thus, x = 1 and x = 2 are the roots of x 2 - 3x + 2 = 0. But how to find them if they are not given? There are different ways of solving quadratic equations.

• Solving quadratic equations by factoring
• Solving quadratic equations by completing the square
• Solving quadratic equations by graphing
• Solving quadratic equations by quadratic formula

Apart from these methods, there are some other methods that are used only in specific cases (when the quadratic equation has missing terms) as explained below.

Solving Quadratic Equations Missing b

In a quadratic equation ax 2 + bx + c = 0, if the term with b is missing then the equation becomes ax 2 + c = 0. This can be solved by taking square root on both sides. The process is explained with examples below.

• x 2 - 4 = 0 ⇒ x 2 = 4 ⇒ x = ±√4 ⇒ x = ± 2 Thus, the roots of the equation are 2 and -2.
• x 2 + 49 = 0 ⇒ x 2 = -49 ⇒ x = ±√(-49) ⇒ x = ± 7i Thus, the roots of the equation are 7i and -7i. (note that these are imaginary (or) complex numbers ).

Solving Quadratic Equations Missing c

In a quadratic equation ax 2 + bx + c = 0, if the term with c is missing then the equation becomes ax 2 + bx = 0. To solve this type of equation, we simply factor x out from the left side, set each of the factors to zero, and solve. The process is explained with examples below.

• x 2 - 5x = 0 ⇒ x (x - 5) = 0 ⇒ x = 0; x - 5 = 0 ⇒ x = 0; x = 5 Thus, the roots of the equation are 0 and 5.
• x 2 + 21x = 0 ⇒ x (x + 21) = 0 ⇒ x = 0; x + 21 = 0 ⇒ x = 0; x = -21 Thus, the roots of the equation are 0 and -11.

Now, we will learn the methods of solving the quadratic equations in each of the above-mentioned methods.

Solving Quadratics by Factoring

Solving quadratics by factoring is one of the famous methods used to solve quadratic equations. The step-by-step process of solving quadratic equations by factoring is explained along with an example.

• Step - 1: Get the equation into standard form. i.e., Get all the terms of to one side (usually to left side) of the equation such that the other side is 0.
• Step - 2: Factor the quadratic expression. If you want to know how to factor a quadratic expression, click here .
• Step - 3: By zero product property , set each of the factors to zero.
• Step - 4: Solve each of the above equations .

Example: Solve the quadratic equation x 2 - 3x + 2 = 0 by factoring it.

Factoring the left side part, we get (x - 1) (x - 2) = 0.

Then x - 1 = 0 (or) x - 2 = 0

which gives x = 1 (or) x = 2.

Thus, the solutions of the quadratic equation x 2 - 3x + 2 = 0 are 1 and 2. This method is applicable only when the quadratic expression is factorable. If it is NOT factorable, then we can use one of the other methods as explained below. Similar to quadratic equations we have solutions for linear equations, which are used to solve linear programming problems.

Solving Quadratic Equations by Completing Square

Completing the square means writing the quadratic expression ax 2 + bx + c into the form a (x - h) 2 + k (which is also known as vertex form ), where h = -b/2a and 'k' can be obtained by substituting x = h in ax 2 + bx + c. The step-by-step process of solving the quadratic equations by completing the square is explained along with an example.

• Step - 1: Get the equation into standard form.
• Step - 2: Complete the square on the left side. If you want to know how to complete the square, click here .
• Step - 3: Solve it for x (We will have to take square root on both sides along the way).

Example: Solve 2x 2 + 8x = -3 by completing the square.

The given equation in the standard form is 2x 2 + 8x + 3 = 0. Completing the square on the left side, we get 2 (x + 2) 2 - 5 = 0. Now solving it for x,

Adding 5 on both sides, 2 (x + 2) 2 = 5 Dividing both sides by 2, (x + 2) 2 = 5/2 Taking square root on both sides, x + 2 = √(5/2) = √5/√2 · √2/√2 = √10/2 Subtracting 2 from both sides, x = -2 ± (√10/2) = (-4 ± √10) / 2

Thus, the roots of the quadratic equation 2x 2 + 8x = -3 are (-4 + √10)/2 and (-4 - √10)/2.

Solving Quadratics by Graphing

For solving the quadratics by graphing , we first have to graph the quadratic expression (when the equation is in the standard form) either manually or by using a graphing calculator. Then the x-intercept (s) of the graph (the point(s) where the graph cuts the x-axis) are nothing but the roots of the quadratic equation. Here are the steps to solve quadratic equations by graphing.

• Step - 1: Get into the standard form.
• Step - 2: Graph the quadratic expression (which is on the left side).
• Step - 3: Identify the x-intercepts.
• Step - 4: The x-coordinates of the x-intercepts are nothing but the roots of the quadratic equation.

Example: Solve the quadratic 3x 2 + 5 = 11x by graphing.

Thus, the solutions of the quadratic equation 3x 2 + 5 = 11x are 0.532 and 3.135.

By seeing the above example, we can see that the graphing method of solving quadratic equations may not give the exact solutions (i.e., it gives only the decimal approximations of the roots if they are irrational ). i.e., if we solve the same equation using completing the square, we get x = (11 + √61) / 6 and x = (11 - √61) / 6. But we cannot get these exact roots by the graphing method.

What if the graph does not intersect the x-axis at all? It means that the quadratic equation has two complex roots. i.e., the graphing method is NOT helpful to find the roots if they are complex numbers. We can use the quadratic formula (which is explained in the next section) to find any type of roots.

Solving Quadratic Equations by Quadratic Formula

As we have already seen, the previous methods for solving the quadratic equations have some limitations such as the factoring method is useful only when the quadratic expression is factorable, the graphing method is useful only when the quadratic equation has real roots, etc. But solving quadratic equations by quadratic formula overcomes all these limitations and is useful to solve any type of quadratic equation. Here is the step-by-step explanation of solving quadratics by quadratic formula.

• Step - 2: Compare the equation with ax 2 + bx + c = 0 and find the values of a, b, and c.
• Step - 3: Substitute the values into the quadratic formula which says x = [-b ± √(b² - 4ac)] / (2a). Then we get
• Step - 4: Simplify.

Example: Solve the quadratic equation 2x 2 = 3x - 5 by the quadratic formula.

The above equation in standard form is 2x 2 - 3x + 5 = 0.

Comparing the equation with ax 2 + bx + c = 0, we get a = 2, b = -3. and c = 5.

Substitute the values into the quadratic formula

x = [-(-3) ± √((-3)² - 4(2)(5))] / (2(2)) = [ 3 ± √(9 - 40) ] / 4 = [ 3 ± √(-31) ] / 4 = [ 3 ± i√(31) ] / 4

Thus, the roots of the quadratic equation 2x 2 = 3x - 5 are [ 3 + i√(31) ] / 4 and [ 3 - i√(31) ] / 4. In the quardratic formula, the expression b² - 4ac is called the discriminant (that is denoted by D). i.e., D = b² - 4ac. This is used to determine the nature of roots of the quadratic equation.

Nature of Roots Using Discriminant

• If D > 0, then the equation ax 2 + bx + c = 0 has two real and distinct roots.
• If D = 0, then the equation ax 2 + bx + c = 0 has only one real root.
• If D < 0, then the equation ax 2 + bx + c = 0 has two distinct complex roots.

Thus, using the discriminant, we can find the number of solutions of quadratic equations without actually solving it.

Important Notes on Solving Quadratic Equations:

• The factoring method cannot be applied when the quadratic expression is NOT factorable.
• The graphing method cannot give the complex roots and also it cannot give the exact roots in case the quadratic equation has irrational roots.
• Completing the square method and quadratic formula method can be applied to solve any type of quadratic equation.
• The roots of the quadratic equation are also known as "solutions" or "zeros".
• For any quadratic equation ax 2 + bx + c = 0, the sum of the roots = -b/a the product of the roots = c/a.

☛Related Topics:

• Solving Quadratic Equations by Quadratic Formula Calculator
• Solving Quadratic Equations by Completing Square Calculator
• Roots of Quadratic Equation Calculator
• Solving Quadratic Equations by Factoring Calculator

Solving Quadratic Equations Examples

Example 1: The length of a park is 5 ft less than twice its width. Find the dimensions of the park if its area is 250 square feet.

Let the width of the park be x ft.

Then the length of the park = (2x - 5) ft.

Its area = 250 ft 2

length × width = 250

(2x - 5) x = 250

2x 2 - 5x - 250 = 0

Hence, this is a word problem related to solving quadratics. Let us solve this quadratic equation by factoring.

Here a = 2, b = -5 and c = -250.

ac = 2(-250) = -500.

Two numbers whose sum is -5 and whose product is -500 are -25 and 20. So we split the middle term using these two numbers.

2x 2 - 25x + 20x - 250 = 0

x (2x - 25) + 10 (2x - 25) = 0

(2x - 25) (x + 10) = 0

2x - 25 = 0 (or) x + 10 = 0

x = 25/2 = 12.5 (or) x = -10

x = 12.5 as x cannot be negative.

So width = 12.5 ft and length = (2x - 5) ft = 2(12.5) - 5 = 20 ft.

Answer: The dimensions of the park are 20 ft × 12.5 ft.

Example 2: If twice the difference of a number and 6 is equal to -2 times its square, then find the number(s).

Let the required number be x. Then

2(x - 6) = -2x 2

Let us solve this quadratic equation by factoring. For this, we have to convert this into standard form . Then

2x 2 + 2x - 12 = 0

Here a = 2, b = 2 and c = -12.

ac = 2(-12) = -24.

Two numbers whose sum is 2 and whose product is -24 are 6 and -4. So we split the middle term using these two numbers.

2x 2 + 6x - 4x - 12 = 0

2x (x + 3) - 4 (x + 3) = 0

(x + 3) (2x - 4) = 0

x + 3 = 0, 2x - 4 = 0

x = -3, x = 2

Answer: The required numbers are -3 and 2.

Example 3: The product of two positive consecutive numbers is 156. Find the two numbers.

Let us assume that the two consecutive numbers be x and x + 1. Then

x (x + 1) = 156

x 2 + x - 156 = 0

Let us solve this quadratic equation by factoring.

Here a = 1, b = 1 and c = -156.

ac = 1(-156) = -156.

Two numbers whose sum is 1 and whose product is -156 are 13 and -12. So we split the middle term using these two numbers.

x 2 + 13x - 12x - 156 = 0

x (x + 13) - 12 (x + 13) = 0

(x + 13) (x - 12) = 0

x + 13 = 0, x - 12 = 0

x = -13 (or) x = 12

Since x is positive (given), x cannot be -13. So x = 12.

Answer: The required consecutive numbers are 12 and 13 (12 + 1).

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Practice Questions on Solving Quadratic Equations

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FAQs on Solving Quadratic Equations

What is the meaning of solving quadratic equations.

Solving quadratic equations means finding its solutions or roots. i.e., it is the process of finding the values of the variable that satisfy the equation.

What are the Most Popular Ways of Solving Quadratic Equations?

There are different ways to solve quadratics. But the most popular ways are "solving quadratic equations by factoring" and "solving quadratic equations by quadratic formula".

What are the Steps in Solving Quadratic Equations by Graphing?

To solve quadratics by graphing , first get into standard form ax 2 + bx + c = 0. Then graph the quadratic expression ax 2 + bx + c. Find where the graph is intersecting the x-axis. The x-coordinate of the x-intercept(s) are nothing but the solutions of the quadratic equation.

What are 4 ways to Solving Quadratics?

There are 4 ways for solving quadratic equations.

• by factoring
• by completing square
• by graphing
• by quadratic formula

How to Solve Quadratic Equations by Quadratic Formula?

The solutions of a quadratic equation ax 2 + bx + c = 0 are given by the quadratic formula x = [-b ± √(b² - 4ac)] / (2a). So to solve a quadratic equation using quadratic formula, just get the equation into standard form ax 2 + bx + c = 0, and apply the quadratic formula.

How do You Know Which Method to Use When Solving Quadratic Equations?

We can solve the quadratic equations of any type using completing the square or the quadratic formula. But if the quadratic expression is factorable, then the factoring method is the easiest to apply. We can solve it by graphing method also, but it gives only approximated real roots (i.e., complex roots cannot be found in this method).

What is the Easiest Way For Solving Quadratic Equations?

The easiest way of solving quadratic equations is the factoring method. But not always quadratic expressions are factorable. In that case, we can either use the quadratic formula or use completing the square method.

What are the Steps in Solving Quadratics by Completing Square?

To solve the quadratic equation ax 2 + bx + c = 0 by completing square, convert ax 2 + bx + c into the vertex form a (x - h) 2 + k where h = -b/2a and k is obtained by substituting x = h in ax 2 + bx + c. Then we can easily solve a (x - h) 2 + k = 0 by isolating x. In this process, we will have to take the square root on both sides.

How to Solve Quadratic Equations by Factoring?

For solving the quadratic equations by factoring, first convert it into the standard form (ax 2 + bx + c = 0). Then factorize the left side part using the techniques of factorizing quadratic expressions, set each of the factors to zero that results in two linear equations, and finally solve the linear equations .

How is the Factored Form Helpful in Solving Quadratic Equations?

If the quadratic expression that is in the standard form of quadratic expression in it is factorable, then we can just set each factor to zero, and solve them. The solutions are nothing but the roots of the quadratic equation.

How to Find the Roots of Quadratic Equations?

The roots of the quadratic equation ax 2 + bx + c = 0 can be found by using the quadratic formula that says x = [-b ± √(b² - 4ac)] / (2a). Also, we can solve them by completing square (or) factoring method (only when they are factorable).

Which Method is Best For Solving Quadratic Equations?

The best method to solve quadratic equations is factoring. But when factoring is not possible, we solve them using the quadratic formula x = [-b ± √(b² - 4ac)] / (2a). If you have a graphing calculator, then graphing method would be the easiest to find the decimal approximation of roots (we cannot find exact roots though).

Free Mathematics Tutorials

Quadratic functions problems with solutions.

quadratic functions problems with detailed solutions are presented along with graphical interpretations of the solutions.

Review Vertex and Discriminant of Quadratic Functions

f(x) = a x 2 + b x + c

If a > 0 , the vertex is a minimum point and the minimum value of the quadratic function f is equal to k. This minimum value occurs at x = h. If a < 0 , the vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This maximum value occurs at x = h. The quadratic function f(x) = a x 2 + b x + c can be written in vertex form as follows:

f(x) = a (x - h) 2 + k

Problems with Solutions

Problem 1 The profit (in thousands of dollars) of a company is given by.

P(x) = 5000 + 1000 x - 5 x 2

• Function P that gives the profit is a quadratic function with the leading coefficient a = - 5. This function (profit) has a maximum value at x = h = - b / (2a) x = h = -1000 / (2(-5)) = 100
• The maximum profit Pmax, when x = 100 thousands is spent on advertising, is given by the maximum value of function P k = c - b 2 / (4 a) b)
• The maximum profit Pmax, when x = 100 thousands is spent on advertising, is also given by P(h = 100) P(100) = 5000 + 1000 (100) - 5 (100) 2 = 55000.
• When the company spends 100 thousands dollars on advertising, the profit is maximum and equals 55000 dollars.

Problem 2 An object is thrown vertically upward with an initial velocity of V o feet/sec. Its distance S(t), in feet, above ground is given by

• S(t) is a quadratic function and the maximum value of S(t)is given by k = c - b 2 /(4 a) = 0 - (v o ) 2 / (4(-16))
• This maximum value of S(t) has to be 300 feet in order for the object to reach a maximum distance above ground of 300 feet. - (v o ) 2 / (4(-16)) = 300
• we now solve - (v o ) 2 / (4(-16)) = 300 for v o v o = 64*300 = 80 √3 feet/sec.

Problem 3 Find the equation of the quadratic function f whose graph passes through the point (2 , -8) and has x intercepts at (1 , 0) and (-2 , 0). Solution to Problem 3

• Since the graph has x intercepts at (1 , 0) and (-2 , 0), the function has zeros at x = 1 and x = - 2 and may be written as follows. f(x) = a (x - 1)(x + 2)
• The graph of f passes through the point (2 , -8), it follows that f(2) = - 8
• which leads to - 8 = a (2 - 1)(2 + 2)
• expand the right side of the above equation and group like terms -8 = 4 a
• Solve the above equation for a to obtain a = - 2
• The equation of f is given by f(x) = - 2 (x - 1)(x + 2)
• Check answer f(1) = 0 f(-2) = 0 f(2) = - 2 (2 - 1)(2 + 2) = -8

Problem 4 Find values of the parameter m so that the graph of the quadratic function f given by

• To find the points of intersection, you need to solve the system of equations y = x 2 + x + 1 y = m x
• Substitute m x for y in the first equation to obtain mx = x 2 + x + 1
• Write the above quadratic equation in standard form. x 2 + x (1 - m) + 1 = 0
• Find the discriminant D of the above equation. D = (1 - m) 2 - 4(1)(1) D = (1 - m) 2 - 4 a)
• For the graph of f and that of the line to have 2 points of intersection, D must be positive, which leads to (1 - m) 2 - 4 > 0
• Solve the above inequality to obtain solution set for m in the intervals (- ∞ , -1) U (3 , + ∞) b)
• For the graph of f and that of the line to have 1 point of intersection, D must be zero, which leads to (1 - m) 2 - 4 = 0
• Solve the above equation to obtain 2 solutions for m. m = -1 m = 3 c)
• For the graph of f and that of the line to have no points of intersection, D must be negative, which leads to (1 - m) 2 - 4 < 0

Problem 5 The quadratic function C(x) = a x 2 + b x + c represents the cost, in thousands of Dollars, of producing x items. C(x) has a minimum value of 120 thousands for x = 2000 and the fixed cost is equal to 200 thousands. Find the coefficients a,b and c. Solution to Problem 5

• Function C is a quadratic function. Its minimum point, which is given as (2000,120) is the vertex of the graph of C. Hence we can write C(x) in vertex form as follows C(x) = a (x - 2000) 2 + 120
• The fixed cost is the value of C(x) when x = 0. Hence C(0) = a (0 - 2000) 2 + 120 = 200
• Solve for a a = 80 / 2000 2 = 0.00002
• We expand C(x) and identify the coefficients a, b and c. C(x) = 0.00002 (x - 2000) 2 + 120 = 0.00002 x 2 - 0.08 x + 200 a = 0.00002 , b = -0.08 and c = 200.

Problem 6 Find the equation of the tangent line to the the graph of f(x) = - x 2 + x - 2 at x = 1. Solution to Problem 6

• There are at least two methods to solve the above question. Method 1
• Let the equation of the tangent line be of the form y = m x + b
• and we therefore need to find m and b. The tangent line passes through the point (1 , f(1)) = (1 , -2)
• Hence the equation in m and b - 2 = m (1) + b or m + b = -2
• To find the point of tangency of the line and the graph of the quadratic function, we need to solve the system y = m x + b and y = - x 2 + x - 2
• Substitute y by m x + b in the second equation of the system to obtain m x + b = - x 2 + x - 2
• Write the above equation in standard form - x 2 + x (1 - m) - 2 - b = 0
• For the line to be tangent to the graph of the quadratic function, the discriminant D of the above equation must be equal to zero. Hence D = b 2 - 4 a c = (1 - m) 2 - 4 (-1) (- 2 - b) = 0
• which gives (1 - (- 2 - b) ) 2 + 4 (- 2 - b) = 0
• Expand, simplify and write the above equation in standard form b 2 2 b + 1 = 0 (b + 1) 2 = 0
• Solve for b b = - 1
• Find m m = - 2 - b = -1

Questions with Solutions

Find the equation of the quadratic function f whose graph has x intercepts at (-1 , 0) and (3 , 0) and a y intercept at (0 , -4).

Question 2 Find values of the parameter c so that the graphs of the quadratic function f given by f(x) = x 2 + x + c and the graph of the line whose equation is given by y = 2 x have: a) 2 points of intersection, b) 1 point of intersection, c) no points of intersection.

Solutions to the Above Questions

Solution to Question 1

• The x intercepts of the graph of f are the zero of f(x). Hence f(x) is of the form f(x) = a (x + 1)(x - 3)
• We now need to find coefficient a using the y intercept f(0) = a(0 + 1)(0 - 3) = - 4
• Solve for a a = 4 / 3
• Hence f(x) = (4 / 3) (x + 1)(x - 3)

Solution to Question 2

• To find the coordinates of the point of intersections of the graphs of f(x) = x 2 + x + c and y = 2 x, we need to solve the system y = x 2 + x + c and y = 2 x
• which by substitution , gives the equation x 2 + x + c = 2x
• Rewrite the above equation in standard form x 2 - x + c = 0
• Find the discriminant D D = 1 - 4 c
• Conclusion If D is positive or c < 1 / 4 , the two graphs intersect at two points. If D is equal to 0 or c = 1 / 4 , the two graphs intersect (touch) at 1 point. If D is negative or c > 1 / 4 , the two graphs have no point of intersection.

More References and Links to Quadratic Functions

• Math Questions With Answers (13): Quadratic Functions.
• Vertex and Intercepts Parabola Problems.
• Find Vertex and Intercepts of Quadratic Functions - Calculator: An applet to solve calculate the vertex and x and y intercepts of the graph of a quadratic function.
• Quadratic functions (general form).
• Quadratic functions (standard form).
• Graphing quadratic functions.
• Solver to Analyze and Graph a Quadratic Function

• Solve equations and inequalities
• Simplify expressions
• Factor polynomials
• Graph equations and inequalities
• Advanced solvers
• All solvers
• Arithmetics
• Determinant
• Percentages
• Scientific Notation
• Inequalities

Quadratic Equations

Solving equations is the central theme of algebra. All skills learned lead eventually to the ability to solve equations and simplify the solutions. In previous chapters we have solved equations of the first degree. You now have the necessary skills to solve equations of the second degree, which are known as quadratic equations .

QUADRATICS SOLVED BY FACTORING

• Identify a quadratic equation.
• Place a quadratic equation in standard form.
• Solve a quadratic equation by factoring.

A quadratic equation is a polynomial equation that contains the second degree, but no higher degree, of the variable.

The standard form of a quadratic equation is ax 2 + bx + c = 0 when a ≠ 0 and a, b, and c are real numbers.

All quadratic equations can be put in standard form, and any equation that can be put in standard form is a quadratic equation. In other words, the standard form represents all quadratic equations.

The solution to an equation is sometimes referred to as the root of the equation.

An important theorem, which cannot be proved at the level of this text, states "Every polynomial equation of degree n has exactly n roots." Using this fact tells us that quadratic equations will always have two solutions. It is possible that the two solutions are equal.

The simplest method of solving quadratics is by factoring. This method cannot always be used, because not all polynomials are factorable, but it is used whenever factoring is possible.

The method of solving by factoring is based on a simple theorem.

If AB = 0, then either A = 0 or B = 0.

We will not attempt to prove this theorem but note carefully what it states. We can never multiply two numbers and obtain an answer of zero unless at least one of the numbers is zero. Of course, both of the numbers can be zero since (0)(0) = 0.

Solution Step 1 Put the equation in standard form.

Step 2 Factor completely.

Step 3 Set each factor equal to zero and solve for x. Since we have (x - 6)(x + 1) = 0, we know that x - 6 = 0 or x + 1 = 0, in which case x = 6 or x = - 1.

Step 4 Check the solution in the original equation. If x = 6, then x 2 - 5x = 6 becomes

Therefore, x = 6 is a solution. If x = - 1, then x 2 - 5x = 6 becomes

Therefore, - 1 is a solution.

The solutions can be indicated either by writing x = 6 and x = - 1 or by using set notation and writing {6, - 1}, which we read "the solution set for x is 6 and - 1." In this text we will use set notation.

Check the solutions in the original equation.

INCOMPLETE QUADRATICS

• Identify an incomplete quadratic equation.
• Solve an incomplete quadratic equation.

If, when an equation is placed in standard form ax 2 + bx + c = 0, either b = 0 or c = 0, the equation is an incomplete quadratic .

5x 2 - 10 = 0 is an incomplete quadratic, since the middle term is missing and therefore b = 0.

When you encounter an incomplete quadratic with c - 0 (third term missing), it can still be solved by factoring.

Notice that if the c term is missing, you can always factor x from the other terms. This means that in all such equations, zero will be one of the solutions. An incomplete quadratic with the b term missing must be solved by another method, since factoring will be possible only in special cases.

Example 3 Solve for x if x 2 - 12 = 0.

Solution Since x 2 - 12 has no common factor and is not the difference of squares, it cannot be factored into rational factors. But, from previous observations, we have the following theorem.

Using this theorem, we have

Note that in this example we have the square of a number equal to a negative number. This can never be true in the real number system and, therefore, we have no real solution.

COMPLETING THE SQUARE

• Identify a perfect square trinomial.
• Complete the third term to make a perfect square trinomial.
• Solve a quadratic equation by completing the square.

From your experience in factoring you already realize that not all polynomials are factorable. Therefore, we need a method for solving quadratics that are not factorable. The method needed is called "completing the square."

First let us review the meaning of "perfect square trinomial." When we square a binomial we obtain a perfect square trinomial. The general form is (a + b) 2 = a 2 + 2ab + b 2 .

• The other term is either plus or minus two times the product of the square roots of the other two terms.

The -7 term immediately says this cannot be a perfect square trinomial. The task in completing the square is to find a number to replace the -7 such that there will be a perfect square.

Consider this problem: Fill in the blank so that "x 2 + 6x + _______" will be a perfect square trinomial. From the two conditions for a perfect square trinomial we know that the blank must contain a perfect square and that 6x must be twice the product of the square root of x 2 and the number in the blank. Since x is already present in 6x and is a square root of x 2 , then 6 must be twice the square root of the number we place in the blank. In other words, if we first take half of 6 and then square that result, we will obtain the necessary number for the blank.

Therefore x 2 + 6x + 9 is a perfect square trinomial.

Now let's consider how we can use completing the square to solve quadratic equations.

Example 5 Solve x 2 + 6x - 7 = 0 by completing the square.

Solution First we notice that the -7 term must be replaced if we are to have a perfect square trinomial, so we will rewrite the equation, leaving a blank for the needed number.

At this point, be careful not to violate any rules of algebra. For instance, note that the second form came from adding +7 to both sides of the equation. Never add something to one side without adding the same thing to the other side.

Now we find half of 6 = 3 and 3 2 = 9, to give us the number for the blank. Again, if we place a 9 in the blank we must also add 9 to the right side as well.

Now factor the perfect square trinomial, which gives

Example 6 Solve 2x 2 + 12x - 4 = 0 by completing the square.

Solution This problem brings in another difficulty. The first term, 2x 2 , is not a perfect square. We will correct this by dividing all terms of the equation by 2 and obtain

We now add 2 to both sides, giving

Example 7 Solve 3x 2 + 7x - 9 = 0 by completing the square.

Solution Step 1 Divide all terms by 3.

Step 2 Rewrite the equation, leaving a blank for the term necessary to complete the square.

Step 3 Find the square of half of the coefficient of x and add to both sides.

Step 4 Factor the completed square.

Step 5 Take the square root of each side of the equation.

Step 6 Solve for x (two values).

Follow the steps in the previous computation and then note especially the last ine. What is the conclusion when the square of a quantity is equal to a negative number? "No real solution."

In summary, to solve a quadratic equation by completing the square, follow this step-by-step method.

Step 1 If the coefficient of x2 is not 1, divide all terms by that coefficient. Step 2 Rewrite the equation in the form of x2 + bx + _______ = c + _______. Step 3 Find the square of one-half of the coefficient of the x term and add this quantity to both sides of the equation. Step 4 Factor the completed square and combine the numbers on the right-hand side of the equation. Step 5 Find the square root of each side of the equation. Step 6 Solve for x and simplify. If step 5 is not possible, then the equation has no real solution.

THE QUADRATIC FORMULA

• Solve the general quadratic equation by completing the square.
• Solve any quadratic equation by using the quadratic formula.

The standard form of a quadratic equation is ax 2 + bx + c = 0. This means that every quadratic equation can be put in this form. In a sense then ax 2 + bx + c = 0 represents all quadratics. If you can solve this equation, you will have the solution to all quadratic equations.

We will solve the general quadratic equation by the method of completing the square.

To use the quadratic formula you must identify a, b, and c. To do this the given equation must always be placed in standard form.

Not every quadratic equation will have a real solution.

There is no real solution since -47 has no real square root.

This solution should now be simplified.

WORD PROBLEMS

• Identify word problems that require a quadratic equation for their solution.
• Solve word problems involving quadratic equations.

Certain types of word problems can be solved by quadratic equations. The process of outlining and setting up the problem is the same as taught in chapter 5, but with problems solved by quadratics you must be very careful to check the solutions in the problem itself. The physical restrictions within the problem can eliminate one or both of the solutions.

Example 1 If the length of a rectangle is 1 unit more than twice the width, and the area is 55 square units, find the length and width.

Solution The formula for the area of a rectangle is Area = Length X Width. Let x = width, 2x + 1 = length.

At this point, you can see that the solution x = -11/2 is not valid since x represents a measurement of the width and negative numbers are not used for such measurements. Therefore, the solution is

width = x = 5, length = 2x + 1 = 11.

Example 3 If a certain integer is subtracted from 6 times its square, the result is 15. Find the integer.

Solution Let x = the integer. Then

Since neither solution is an integer, the problem has no solution.

Example 4 A farm manager has 200 meters of fence on hand and wishes to enclose a rectangular field so that it will contain 2,400 square meters in area. What should the dimensions of the field be?

Solution Here there are two formulas involved. P = 2l + 2w for the perimeter and A = lw for the area. First using P = 2l + 2w, we get

We can now use the formula A = lw and substitute (100 - l) for w, giving

The field must be 40 meters wide by 60 meters long.

Note that in this problem we actually use a system of equations

P = 2 l + 2 w A = l w.

In general, a system of equations in which a quadratic is involved will be solved by the substitution method. (See chapter 6.)

• A quadratic equation is a polynomial equation in one unknown that contains the second degree, but no higher degree, of the variable.
• The standard form of a quadratic equation is ax 2 + bx + c = 0, when a ≠ 0.
• An incomplete quadratic equation is of the form ax 2 + bx + c = 0, and either b = 0 or c = 0.

• The most direct and generally easiest method of finding the solutions to a quadratic equation is factoring. This method is based on the theorem: if AB = 0, then A = 0 or B = 0. To use this theorem we put the equation in standard form, factor, and set each factor equal to zero.
• To solve a quadratic equation by completing the square, follow these steps: Step 1 If the coefficient of x 2 is not 1, divide all terms by that coefficient. Step 2 Rewrite the equation in the form of x 2 + bx +_____ = c + _____ Step 3 Find the square of one-half of the coefficient of the x term and add this quantity to both sides of the equation. Step 4 Factor the completed square and combine the numbers on the right-hand side of the equation. Step 5 Find the square root of each side of the equation. Step 6 Solve for x and simplify.
• The method of completing the square is used to derive the quadratic formula.
• To use the quadratic formula write the equation in standard form, identify a, b, and c, and substitute these values into the formula. All solutions should be simplified.

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We can help you solve an equation of the form " ax 2 + bx + c = 0 " Just enter the values of a, b and c below :

Is it Quadratic?

Only if it can be put in the form ax 2 + bx + c = 0 , and a is not zero .

The name comes from "quad" meaning square, as the variable is squared (in other words x 2 ).

These are all quadratic equations in disguise:

How Does this Work?

The solution(s) to a quadratic equation can be calculated using the Quadratic Formula :

The "±" means we need to do a plus AND a minus, so there are normally TWO solutions !

The blue part ( b 2 - 4ac ) is called the "discriminant", because it can "discriminate" between the possible types of answer:

• when it is positive, we get two real solutions,
• when it is zero we get just ONE solution,
• when it is negative we get complex solutions.

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Example: 3x^2-2x-1=0

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Number line.

• x^4-5x^2+4=0
• \sqrt{x-1}-x=-7
• \left|3x+1\right|=4
• \log _2(x+1)=\log _3(27)
• 3^x=9^{x+5}
• What is the completing square method?
• Completing the square method is a technique for find the solutions of a quadratic equation of the form ax^2 + bx + c = 0. This method involves completing the square of the quadratic expression to the form (x + d)^2 = e, where d and e are constants.
• What is the golden rule for solving equations?
• The golden rule for solving equations is to keep both sides of the equation balanced so that they are always equal.
• How do you simplify equations?
• To simplify equations, combine like terms, remove parethesis, use the order of operations.
• How do you solve linear equations?
• To solve a linear equation, get the variable on one side of the equation by using inverse operations.

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• High School Math Solutions – Quadratic Equations Calculator, Part 1 A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c...

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