solving quadratic equations homework 5 complex numbers

Solving Quadratic Equations: Worksheets with Answers

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Unit 5: Complex Numbers

Why are we studying this.

In this unit of study, students will continue to deepen their knowledge and understanding of quadratic equations. Students will learn how to solve quadratic equations with complex (non-real) solutions. Students will also learn the concept of imaginary and complex numbers.

Unit Schedule and Assignments

Weekly online, notebook: unit 5, table of contents & pages, hw #12: due 1/11, hw #14: due 1/25, hw #13: due 1/18.

Daily Homework Assignments (w/Answer Keys)

Intro. to imaginary numbers - practice page.

Homework - 1/7 and 1/8

11.2: Operations with Complex Numbers (#1-30, first page ONLY)

Homework - 1/9 and 1/10

11.3: #3-12 ALL (pg.554-555)

Homework - 1/11 and 1/14

11.2: Operations with Complex Numbers (#1-26, second page ONLY)

Unit assessments.

Unit 5 Assessment: Thursday, January 17th (Per. 1 & 3) and Friday, January 18th (Per. 2)

Imaginary Numbers and Complex Numbers

Notes Page - Imaginary Numbers

Notes Page - Complex Numbers

In Class Practice

In Class Practice (Answer Key)

11.2: Operations with Complex Numbers

Class Slides (outline used to make video lesson)

Video Lesson

Practice Packet (w/answers)

Tutorial Videos: +/-/x with Complex Numbers

11.3: finding complex solutions of quadratic equations.

11.3: Textbook Pages

In Class Slides

Notes Page (examples done in class)

UNIT REVIEW RESOURCES

Scavenger Hunt

Triples Activity

Bingo: Operations with Complex Numbers

Unit Review: Slides

Quadratic Equations & Complex Numbers (Algebra 2 - Unit 4) | All Things Algebra®

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What educators are saying

Also included in.

Description

This Quadratic Equations & Complex Numbers Unit Bundle includes guided notes, homework assignments, three quizzes, a study guide and a unit test that cover the following topics:

• Roots of a Quadratic Equation; Solving Quadratics by Graphing

• Factoring Review

• Solving Quadratics by Factoring

• Factored Form/Vertex Form/Standard Form of a Quadratic Equation

• Simplifying Radicals Review

• Solving Quadratics by Square Roots

• Imaginary Numbers

• Solving Square Roots Problems with Imaginary Solutions

• Operations with Complex Numbers

• Organizing the Real and Complex Numbers

• Properties with Complex Numbers

• Solving Quadratics by Completing the Square

• Solving Quadratics by The Quadratic Formula

• The Discriminant

• Review of all Methods; Choosing the Best Method

• Geometric and Consecutive Integer Applications

• Projectile Motion

• Quadratic Regression

• Solving Nonlinear Systems of Equations (Linear-Quadratic and Quadratic-Quadratic) graphically • Solving Nonlinear Systems of Equations (Linear-Quadratic and Quadratic-Quadratic) algebraically

ADDITIONAL COMPONENTS INCLUDED:

(1) Links to Instructional Videos: Links to videos of each lesson in the unit are included. Videos were created by fellow teachers for their students using the guided notes and shared in March 2020 when schools closed with no notice.  Please watch through first before sharing with your students. Many teachers still use these in emergency substitute situations. (2) Editable Assessments: Editable versions of each quiz and the unit test are included. PowerPoint is required to edit these files. Individual problems can be changed to create multiple versions of the assessment. The layout of the assessment itself is not editable. If your Equation Editor is incompatible with mine (I use MathType), simply delete my equation and insert your own.

(3) Google Slides Version of the PDF: The second page of the Video links document contains a link to a Google Slides version of the PDF. Each page is set to the background in Google Slides. There are no text boxes;  this is the PDF in Google Slides.  I am unable to do text boxes at this time but hope this saves you a step if you wish to use it in Slides instead! 

This resource is included in the following bundle(s):

Algebra 2 Curriculum

More Algebra 2 Units:

Unit 1 – Equations and Inequalities

Unit 2 – Linear Functions and Systems

Unit 3 – Parent Functions and Transformations

Unit 5 – Polynomial Functions

Unit 6 – Radical Functions

Unit 7 – Exponential and Logarithmic Functions

Unit 8 – Rational Functions

Unit 9 – Conic Sections

Unit 10 – Sequences and Series

Unit 11 – Probability and Statistics

Unit 12 – Trigonometry

LICENSING TERMS: This purchase includes a license for one teacher only for personal use in their classroom. Licenses are non-transferable , meaning they can not be passed from one teacher to another. No part of this resource is to be shared with colleagues or used by an entire grade level, school, or district without purchasing the proper number of licenses. If you are a coach, principal, or district interested in transferable licenses to accommodate yearly staff changes, please contact me for a quote at [email protected].

COPYRIGHT TERMS: This resource may not be uploaded to the internet in any form, including classroom/personal websites or network drives, unless the site is password protected and can only be accessed by students.

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NCERT Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations describe the properties of complex numbers and their use in finding the roots of quadratic equations. Students are familiar with finding the roots of a quadratic equation in the real numbers set with a non-negative discriminant. In this chapter, they will learn how to solve a quadratic equation with a negative discriminant value. The knowledge of Complex Numbers and Quadratic Equations is particularly useful in advanced calculus. These equations are widely applied in solving algebraic problems. The geometric graph of these equations represents a parabola that doesn’t intersect the x-axis on a quadratic plane. NCERT Solutions Class 11 Maths Chapter 5 will offer the right foundational skills required by students to study these equations and their applications.

Gaining a gradual understanding of complex numbers and operations performed on them is vital for understanding complex roots. The well-explained step-by-step illustrative format of these solutions helps to implement a thorough practice of all topics. Several concepts, formulas, theorems, and problems are explained in these solutions to provide a clear understanding of complex numbers and quadratic equations. With the regular revision of Class 11 Maths NCERT Solutions Chapter 5, students will quickly master how to solve these equations. To learn and practice with NCERT Solutions Chapter 5 Complex Numbers and Quadratic Equations, download the exercise-wise solutions provided in the links below.

• NCERT Solutions Class 11 Maths Chapter 5 Ex 5.1
• NCERT Solutions Class 11 Maths Chapter 5 Ex 5.2
• NCERT Solutions Class 11 Maths Chapter 5 Ex 5.3
• NCERT Solutions Class 11 Maths Chapter 5 Miscellaneous Ex

NCERT Solutions for Class 11 Maths Chapter 5 PDF

NCERT Solutions for Class 11 Maths Chapter 5 is geared towards easy and proficient math learning. The self-explanatory composition of exercises, examples, and illustrations present in these solutions are highly reliable to study the complete syllabus of this chapter. These resources are a highly reliable means of gaining clear and extensive knowledge of complex numbers and their use in algebra . To learn and practice Complex Numbers and Quadratic Equations with these solutions, click on the pdf file links given below.

☛ Download Class 11 Maths NCERT Solutions Chapter 5

NCERT Class 11 Maths Chapter 5   Download PDF

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

NCERT solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are advantageous for promoting students’ fundamental algebraic skills. These well-researched resources offer systematic learning of complex concepts and topics related to them. The considerate blend of illustrations and sample questions incorporated in these solutions is adequate to form the core basics required for solving quadratic equations. With the practice of these solutions, it becomes quite convenient to prepare and revise for exams. To practice the exercise-wise NCERT Solutions Class 11 Maths Complex Numbers and Quadratic Equations, try the links given below.

• Class 11 Maths Chapter 5 Ex 5.1 - 14 Questions
• Class 11 Maths Chapter 5 Ex 5.2 - 8 Questions
• Class 11 Maths Chapter 5 Ex 5.3 - 9 Questions
• Class 11 Maths Chapter 5 Miscellaneous Ex - 20 Questions

☛  Download Class 11 Maths Chapter 5 NCERT Book

Topics Covered: The important topics included in NCERT solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are an introduction to complex numbers , algebra of complex numbers, modulus , and conjugate of a complex number . These solutions comprise important formulas and theorems based on the subtopics like the addition of complex numbers, difference of complex numbers, multiplication of complex numbers , division of two complex numbers , power of i, and square roots of a negative real number identities.

Total Questions: Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations has 31 questions in 3 exercises plus 20 questions in a miscellaneous exercise. These questions are majorly aligned towards imparting a deep understanding of complex numbers, quadratic equations , their applications, theorems, and formulas.

List of Formulas in NCERT Solutions Class 11 Maths Chapter 5

Formulas are an indispensable part of algebra, and therefore, it is vital to memorize formulas for scoring better in exams. Learning formulas, important terms, and concepts related to Complex Numbers and Quadratic Equations will also enable students to form a firm foundation required for mastering these topics. NCERT Solutions Class 11 Maths Chapter 5 is a valuable aid in gaining a detailed knowledge of all the formulas and concepts. These are well-explained for students to imbibe them better. Some of the important terms, formulas, and concepts related to Complex Numbers and Quadratic Equations explained in these solutions are listed below:

• Complex Numbers: Complex Numbers are the numbers of the form a + ib, where a and b are real numbers.
• Addition of Complex Numbers (a+bi) + (c+di) = (a+c) + (b+d)i
• Subtraction of Complex Numbers (a+bi) − (c+di) = (a−c) + (b−d)i
• Multiplication of Complex Numbers (a+bi) × (c+di) = (ac−bd) + (ad+bc)i
• Multiplication Conjugates (a+bi)(a+bi) = a 2 + b 2
• Division of Complex Numbers

(a+bi)/(c+di) = (a+bi/c+di) × (c−di) / (c−di) = (ac+bd / c 2 +d 2 ) + (bc−ad / c 2 +d 2 )i

• Equality of Complex Numbers Formula

a+bi = c+di ⇔ a = c and b = d

• Fundamental theorem of Algebra: The fundamental theorem of algebra states that every polynomial equation having complex coefficients and degree >=1 has at least one complex root. This theorem also holds true for the equations with real coefficients, as real numbers are also complex numbers with an imaginary part.
• Roots of Quadratic Equation: The roots of a quadratic equation ax² + bx + c = 0 are given by the formula:

x = [-b±√(b 2 -4ac)]/2a

FAQs on NCERT Solutions Class 11 Maths Chapter 5

What is the importance of ncert solutions for class 11 maths chapter 5 complex numbers and quadratic equations.

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are the most accurate and reliable means of math learning. These well-researched solutions written by math experts are highly prominent to encourage practice and confidence in students. The exercises provided in these solutions are apt to understand complex numbers concepts and their use in quadratic equations. With the regular practice of these solutions, it becomes pretty convenient to score well in exams.

What are the Important Topics Covered in Class 11 Maths NCERT Solutions Chapter 5?

NCERT Solutions Class 11 Maths Chapter 5 starts with an introduction to complex numbers, algebra of complex numbers, and their polar representation. The subtopics covered under these topics are the sum of complex numbers, differences of complex numbers, product, division of complex numbers, power of i, square roots of a negative real number, modulus, and conjugate of complex numbers . The other important topics covered in this chapter are based on solving quadratic equations with complex roots and their fundamental theorems.

Do I Need to Practice all Questions Provided in NCERT Solutions Class 11 Maths Complex Numbers and Quadratic Equations?

NCERT Solutions Class 11 Maths Complex Numbers and Quadratic Equations are well-curated for students to learn complex numbers with ease. Every question included in these solutions provides conceptual clarity for kids to master them easily. The fundamental knowledge of Complex Numbers and Quadratic Equations, as well as their use, will allow children to apply their knowledge to practical situations. These solutions also form the basis for studying advanced math topics, including calculus .

How Many Questions are there in Class 11 Maths NCERT Solutions Chapter 5 Complex Numbers and Quadratic Equations?

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations has a total of 51 questions in 4 exercises. Solving and practicing these sums will help students gain a deep knowledge of important formulas, theorems, and arithmetic operations related to Complex Numbers and Quadratic Equations. The simplistic math vocabulary used in these solutions is quite proficient in easily imparting a clear and step-by-step understanding of each topic.

What are the Important Formulas in NCERT Solutions Class 11 Maths Chapter 5?

NCERT Solutions Class 11 Maths Chapter 5 explains the formulas related to the sum and difference of two angles in Complex Numbers and Quadratic Equations. Some of the important concepts and formulas related to this topic are based on the arithmetic of complex numbers, fundamental theorem of algebra, and quadratic equations. These concepts are elaborated in detail with the help of engaging illustrations and examples.

Why Should I Practice NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations?

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations is a comprehensive guide that promotes the accurate and precise knowledge of complex numbers and their use in quadratic equations. With the practice of these solutions, students can learn to excel in exams. Students can cover all the topics provided in the CBSE Class 11 Maths Chapter 5 by using these resources. The pattern of problems incorporated in these solutions is pretty amiable to deliver simple and easy learning.

• NCERT Solutions
• NCERT Class 11
• NCERT 11 Maths
• Chapter 5: Complex Numbers And Quadratic Equations

NCERT Solutions For Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Ncert solutions class 11 maths chapter 5 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 4.

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are prepared by the expert teachers at BYJU’S. These NCERT Solutions of Maths help students in solving problems quickly, accurately and efficiently. Also, BYJU’S provides step-by-step solutions for all NCERT problems, thereby ensuring students understand them and clear their board exams with flying colours. The chapter Complex Numbers and Quadratic Equations is categorised under the CBSE Syllabus for 2023-24 and includes different critical Mathematical theorems and formulae. The NCERT textbook has many practice problems to cover all these concepts, which would help students easily understand higher concepts in future. BYJU’S provides solutions for all these problems with proper explanations. These NCERT Solutions from BYJU’S help students who aim to clear their exams even with last-minute preparations. However, NCERT Solutions for Class 11 Maths are focused on mastering the concepts along with gaining broader knowledge.

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Access answers of maths ncert class 11 chapter 5 – complex numbers and quadratic equations.

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Access the exercises of Maths NCERT Class 11 Chapter 5

Exercise 5.1 Solutions 14 Questions Exercise 5.2 Solutions 8 Questions Exercise 5.3 Solutions 10 Questions Miscellaneous Exercise on Chapter 5 Solutions 20 Questions; the summarisation of the topics discussed in Chapter 5 of the Class 11 NCERT curriculum is listed below.

Access NCERT Solutions for Class 11 Maths Chapter 5

Exercise 5.1 Page No: 103

Express each of the complex numbers given in Exercises 1 to 10 in the form a + ib.

1. (5i) (-3/5i)

(5i) (-3/5i) = 5 x (-3/5) x i 2

(5i) (-3/5i) = 3 + i0

2. i 9 + i 19

i 9 + i 19 = (i 2 ) 4 . i + (i 2 ) 9 . i

= (-1) 4 . i + (-1) 9 .i

= 1 x i + -1 x i

i 9 + i 19 = 0 + i0

Now, multiplying the numerator and denominator by i we get

i -39 = 1 x i / (-i x i)

i -39 = 0 + i

4. 3(7 +  i 7) +  i (7 +  i 7)

3(7 +  i 7) +  i (7 +  i 7) = 21 + i 21 + i 7 + i 2 7

= 14 + i 28

3(7 +  i 7) +  i (7 +  i 7) = 14 + i 28

5. (1 –  i ) – (–1 +  i 6)

(1 –  i ) – (–1 +  i 6) = 1 –  i + 1 –  i 6

(1 –  i ) – (–1 +  i 6) = 2 – i 7

8. (1 –  i ) 4

(1 –  i ) 4 = [(1 –  i ) 2 ] 2

= [1 + i 2 – 2 i ] 2

Hence, (1 –  i ) 4 = -4 + 0 i

9. (1/3 + 3 i ) 3

Hence, (1/3 + 3 i ) 3 = -242/27 – 26 i

10. (-2 – 1/3 i ) 3

(-2 – 1/3 i ) 3 = -22/3 – 107/27 i

Find the multiplicative inverse of each of the complex numbers given in Exercises 11 to 13.

Let’s consider z  = 4 – 3 i

= 4 + 3 i  and

|z| 2 = 4 2 + (-3) 2 = 16 + 9 = 25

Thus, the multiplicative inverse of 4 – 3 i  is given by z -1

12. √5 + 3 i

Let’s consider  z  = √5 + 3 i

|z| 2 = (√5) 2 + 3 2 = 5 + 9 = 14

Thus, the multiplicative inverse of √5 + 3 i is given by z -1

Let’s consider  z  = – i

Thus, the multiplicative inverse of – i  is given by z -1

14. Express the following expression in the form of a + ib:

Exercise 5.2 Page No: 108

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

1. z = – 1 – i √3

2. z = -√3 + i

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

Exercise 5.3 Page No: 109

Solve each of the following equations:

1. x 2 + 3 = 0

Given the quadratic equation,

x 2  + 3 = 0

On comparing it with  ax 2  +  bx  +  c  = 0, we have

a  = 1,  b  = 0, and  c  = 3

So, the discriminant of the given equation will be

D =  b 2  – 4 ac  = 0 2  – 4 × 1 × 3 = –12

Hence, the required solutions are

2. 2x 2 + x + 1 = 0

2 x 2  +  x  + 1 = 0

a  = 2,  b  = 1, and  c  = 1

D =  b 2  – 4 ac  = 1 2  – 4 × 2 × 1 = 1 – 8 = –7

3. x 2 + 3x + 9 = 0

x 2  + 3 x  + 9 = 0

a  = 1,  b  = 3, and  c  = 9

D =  b 2  – 4 ac  = 3 2  – 4 × 1 × 9 = 9 – 36 = –27

4. – x 2  +  x  – 2 = 0

– x 2  +  x  – 2 = 0

a  = –1,  b  = 1, and  c  = –2

D =  b 2  – 4 ac  = 1 2  – 4 × (–1) × (–2) = 1 – 8 = –7

5. x 2  + 3 x  + 5 = 0

x 2  + 3 x  + 5 = 0

a  = 1,  b  = 3, and  c  = 5

D =  b 2  – 4 ac  = 3 2  – 4 × 1 × 5 =9 – 20 = –11

6. x 2  –  x  + 2 = 0

x 2  –  x  + 2 = 0

a  = 1,  b  = –1, and  c  = 2

So, the discriminant of the given equation is

D =  b 2  – 4 ac  = (–1) 2  – 4 × 1 × 2 = 1 – 8 = –7

7. √2 x 2  + x  + √2 = 0

√2 x 2  + x  + √2 = 0

a  = √2,  b  = 1, and  c  = √2

D =  b 2  – 4 ac  = (1) 2  – 4 × √2 × √2 = 1 – 8 = –7

8. √3 x 2  – √2 x  + 3√3 = 0

√3 x 2  – √2 x  + 3√3 = 0

a  = √3,  b  = -√2, and  c  = 3√3

D =  b 2  – 4 ac  = (-√2) 2  – 4 × √3 × 3√3 = 2 – 36 = –34

9. x 2  + x  + 1/√2 = 0

x 2  + x  + 1/√2 = 0

It can be rewritten as,

√2 x 2  + √2 x  + 1 = 0

a  = √2,  b  = √2, and  c  = 1

D =  b 2  – 4 ac  = (√2) 2  – 4 × √2 × 1 = 2 – 4√2 = 2(1 – 2√2)

10. x 2  + x /√2 + 1 = 0

x 2  + x /√2 + 1 = 0

D =  b 2  – 4 ac  = (1) 2  – 4 × √2 × √2 = 1 – 8 = -7

Miscellaneous Exercise Page No: 112

2. For any two complex numbers   z 1  and z 2 , prove that

Re (z 1 z 2 )   = Re z 1  Re z 2  – Im z 1  Im z 2

3. Reduce to the standard form.

5. Convert the following into the polar form:

Solve each of the equations in Exercises 6 to 9.

6. 3x 2 – 4x + 20/3 = 0

Given the quadratic equation, 3x 2 – 4x + 20/3 = 0

It can be re-written as: 9x 2 – 12x + 20 = 0

On comparing it with  ax 2  +  bx  +  c  = 0, we get

a  = 9,  b  = –12, and  c  = 20

D =  b 2  – 4 ac  = (–12) 2  – 4 × 9 × 20 = 144 – 720 = –576

7. x 2 – 2x + 3/2 = 0

Given the quadratic equation, x 2 – 2x + 3/2 = 0

It can be re-written as 2x 2 – 4x + 3 = 0

a  = 2,  b  = –4, and  c  = 3

D =  b 2  – 4 ac  = (–4) 2  – 4 × 2 × 3 = 16 – 24 = –8

8. 27x 2 – 10x + 1 = 0

Given the quadratic equation, 27 x 2  – 10 x  + 1 = 0

a  = 27,  b  = –10, and  c  = 1

D =  b 2  – 4 ac  = (–10) 2  – 4 × 27 × 1 = 100 – 108 = –8

9. 21x 2 – 28x + 10 = 0

Given the quadratic equation, 21 x 2  – 28 x  + 10 = 0

On comparing it with  ax 2  +  bx  +  c  = 0, we have

a  = 21,  b  = –28, and  c  = 10

D =  b 2  – 4 ac  = (–28) 2  – 4 × 21 × 10 = 784 – 840 = –56

10. If z 1 = 2 – i , z 2 = 1 + i , find

Given, z 1 = 2 – i , z 2 = 1 + i

12. Let z 1 = 2 – i , z 2 = -2 + i . Find

13. Find the modulus and argument of the complex number.

14. Find the real numbers  x  and  y  if ( x  –  iy ) (3 + 5 i ) is the conjugate of – 6 – 24 i .

Let’s assume z = ( x  –  iy ) (3 + 5 i )

(3x + 5y) – i (5x – 3y) = -6 -24 i

On equating real and imaginary parts, we have

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

Performing (i) x 3 + (ii) x 5, we get

(9x + 15y) + (25x – 15y) = -18 + 120

x = 102/34 = 3

Putting the value of  x  in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

Therefore, the values of  x  and  y are 3 and –3, respectively.

15. Find the modulus of

16. If ( x  +  iy ) 3  =  u  +  iv , then show that

17. If α and β are different complex numbers with |β| = 1, then find

18. Find the number of non-zero integral solutions of the equation |1 – i| x = 2 x.

Therefore, 0 is the only integral solution of the given equation.

Hence, the number of non-zero integral solutions of the given equation is 0.

19. If ( a  +  ib ) ( c  +  id ) ( e  +  if ) ( g  +  ih ) = A +  i B, then show that

( a 2  +  b 2 ) ( c 2  +  d 2 ) ( e 2  +  f 2 ) ( g 2  +  h 2 ) = A 2  + B 2

20. If, then find the least positive integral value of  m .

Thus, the least positive integer is 1.

Therefore, the least positive integral value of  m  is 4 (= 4 × 1).

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations

Chapter 5 of Class 11 Complex Numbers and Quadratic Equations has 3 exercises and a miscellaneous exercise to help the students practise the required number of problems to understand all the concepts. The topics and sub-topics discussed in the PDF of NCERT Solutions for Class 11 of this chapter include 5.1 Introduction We know that some of the quadratic equations have no real solutions. That means the solution of such equations includes complex numbers. Here, we have found the solution of a quadratic equation ax 2 + bx + c = 0 where D = b 2 – 4ac < 0. 5.2 Complex Numbers Definition of complex numbers, examples and explanations about the real and imaginary parts of complex numbers have been discussed in this section. Class 11 Maths NCERT Supplementary Exercise Solutions PDF helps the students to understand the questions in detail. 5.3 Algebra of Complex Numbers 5.3.1 Addition of two complex numbers

5.3.2 Difference of two complex numbers

5.3.3 Multiplication of two complex numbers

5.3.4 Division of two complex number

5.3.5 Power of i

5.3.6 The square roots of a negative real number

5.3.7 Identities

After studying these exercises, students are able to understand the basic BODMAS operations on complex numbers, along with their properties, power of i, square root of a negative real number and identities of complex numbers. 5.4 The Modulus and the Conjugate of a Complex Number The detailed explanation provides the modulus and conjugate of a complex number with solved examples. 5.5 Argand Plane and Polar Representation 5.5.1 Polar representation of a complex number

In this section, it has been explained how to write the ordered pairs for the given complex numbers, the definition of a Complex plane or Argand plane and the polar representation of the ordered pairs in terms of complex numbers.

• A number of the form a + ib, where a and b are real numbers, is called a complex number, “ a” is called the real part, and “ b” is called the imaginary part of the complex number
• z 1 + z 2 = (a + c) + i (b + d)
• z 1 z 2 = (ac – bd) + i (ad + bc)
• For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists a complex number, denoted by 1/z or z–1, called the multiplicative inverse of z
• For any integer k, i 4k = 1, i 4k + 1 = i, i 4k + 2 = – 1, i 4k + 3 = – i
• The polar form of the complex number z = x + iy is r (cosθ + i sinθ)
• A polynomial equation of n degree has n roots.

Disclaimer – 

Dropped Topics – 

5.5.1 Polar Representation of a Complex Number 5.6 Quadratic Equation Example 11 and Exercise 5.3 Examples 13, 15, 16 Ques. 5–8, 9 and 13 (Miscellaneous Exercise) Last three points in the Summary 5.7 Square-root of a Complex Number

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 5

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NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Complex Numbers and Quadratic Equations Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations All Exercises were prepared by Experienced LearnCBSE.in Teachers.

Free download NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1, Ex 5.2, Ex 5.3 and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

• Complex Numbers and Quadratic Equations Class 11 Ex 5.1
• Complex Numbers and Quadratic Equations Class 11 Ex 5.2
• Complex Numbers and Quadratic Equations Class 11 Ex 5.3
• Complex Numbers and Quadratic Equations Class 11 Miscellaneous Exercise
• सम्मिश्र संख्याएँ और द्विघातीय समीकरण प्रश्नावली 5.1 का हल हिंदी में
• सम्मिश्र संख्याएँ और द्विघातीय समीकरण प्रश्नावली 5.2 का हल हिंदी में
• सम्मिश्र संख्याएँ और द्विघातीय समीकरण प्रश्नावली 5.3 का हल हिंदी में
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• Complex Numbers and Quadratic Equations Class 11 Notes
• Complex Numbers and Quadratic Equations NCERT Exemplar
• RD Sharma Class 11 Maths Complex Numbers
• RD Sharma Class 11 Maths Quadratic Equations
• JEE Main Mathematics Complex Numbers
• JEE Main Mathematics Quadratic Equations And Expressions

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NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations (सम्मिश्र संख्याएँ और द्विघातीय समीकरण) Hindi medium Ex 5.1

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• Chapter 1 Sets
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• Chapter 12 Introduction to Three Dimensional Geometry
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How to Solve Quadratic Equations with Complex Numbers

When solving quadratic equations using the quadratic formula , you sometimes get a negative value under the square root. In these cases, the equation does not have any real solutions.

But now that you’re working with complex numbers , you’re able to find all the solutions to quadratic equations. The reason for this is the fact that the imaginary unit i can be utilized to find complex solutions to the quadratic formula.

Complex Quadratic Equations

Let a , b , c ∈ ℂ be complex numbers with a ≠ 0 . Then a z 2 + b z + c = 0 has the following solutions:

If the expression b 2 − 4 a c is negative, you have to use the imaginary unit i to find the solutions.

Solve z 2 − 4 z + 5 = 0 for z

You recognize the coefficients of the equation to be a = 1 , b = − 4 and c = 5 . This means that you can use the quadratic formula to find the solutions:

Since you have a negative number inside the square root, the equation has no real solutions. However, by utilizing the imaginary, you can still find complex solutions:

Thus the final solutions to the equation are:

Note! In Example 1 you used the following relation:

This relation does not hold in general for complex numbers. Using this relation for complex numbers can yield inconsistencies. The following is an example of such inconsistencies:

The reason this calculation yields a contradiction is that − 1 ⋅ − 1 ≠ − 1 ⋅ − 1 . The rule stating that a b = a b only holds when both a and b are positive numbers.

By using the quadratic formula, you can even solve quadratic equations involving complex coefficients. In those cases, you need some knowledge about complex roots in order to simplify the expression inside the square root.

Solve 1 2 z 2 + i z + 3 2 i = 0 for z

You recognize the coefficients as a = 1 2 , b = i and c = 3 2 i . This means that you can use the quadratic formula to find the solutions:

All complex numbers w have two square roots. In this case you only need to consider the root whose argument lies in the interval [ 0 , π ) . The reason for this is that both solutions are included in ± w .

In order to find the square root of w = − 1 − 3 i , you first need to write w in polar form . In this case, the norm of w is r = 2 , while the argument is 𝜃 = 4 π 3 . Thus w = 2 e i 4 π 3 in polar form. You can now find the square root of w by taking the square root of the norm of w and dividing the argument of w by 2 . Thus the square root of w is 2 e i 2 π 3 . In Cartesian form, the square root is written as − 2 2 + 6 2 i . Thus the solutions to the equation are:

2.5 Quadratic Equations

Learning objectives.

In this section, you will:

• Solve quadratic equations by factoring.
• Solve quadratic equations by the square root property.
• Solve quadratic equations by completing the square.
• Solve quadratic equations by using the quadratic formula.

The computer monitor on the left in Figure 1 is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.

• Solving Quadratic Equations by Factoring

An equation containing a second-degree polynomial is called a quadratic equation . For example, equations such as 2 x 2 + 3 x − 1 = 0 2 x 2 + 3 x − 1 = 0 and x 2 − 4 = 0 x 2 − 4 = 0 are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.

Often the easiest method of solving a quadratic equation is factoring . Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.

If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if a ⋅ b = 0 , a ⋅ b = 0 , then a = 0 a = 0 or b = 0 , b = 0 , where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.

Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression ( x − 2 ) ( x + 3 ) ( x − 2 ) ( x + 3 ) by multiplying the two factors together.

The product is a quadratic expression. Set equal to zero, x 2 + x − 6 = 0 x 2 + x − 6 = 0 is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.

The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , where a , b , and c are real numbers, and a ≠ 0. a ≠ 0. The equation x 2 + x − 6 = 0 x 2 + x − 6 = 0 is in standard form.

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.

The Zero-Product Property and Quadratic Equations

The zero-product property states

where a and b are real numbers or algebraic expressions.

A quadratic equation is an equation containing a second-degree polynomial; for example

where a , b , and c are real numbers, and if a ≠ 0 , a ≠ 0 , it is in standard form.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation x 2 + x − 6 = 0 , x 2 + x − 6 = 0 , the leading coefficient, or the coefficient of x 2 , x 2 , is 1. We have one method of factoring quadratic equations in this form.

Given a quadratic equation with the leading coefficient of 1, factor it.

• Find two numbers whose product equals c and whose sum equals b .
• Use those numbers to write two factors of the form ( x + k ) or  ( x − k ) , ( x + k ) or  ( x − k ) , where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and −2 , −2 , the factors are ( x + 1 ) ( x − 2 ) . ( x + 1 ) ( x − 2 ) .
• Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Factoring and Solving a Quadratic with Leading Coefficient of 1

Factor and solve the equation: x 2 + x − 6 = 0. x 2 + x − 6 = 0.

To factor x 2 + x − 6 = 0 , x 2 + x − 6 = 0 , we look for two numbers whose product equals −6 −6 and whose sum equals 1. Begin by looking at the possible factors of −6. −6.

The last pair, 3 ⋅ ( −2 ) 3 ⋅ ( −2 ) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

The two solutions are 2 2 and −3. −3. We can see how the solutions relate to the graph in Figure 2 . The solutions are the x- intercepts of y = x 2 + x − 6 = 0. y = x 2 + x − 6 = 0.

Factor and solve the quadratic equation: x 2 − 5 x − 6 = 0. x 2 − 5 x − 6 = 0.

Solve the Quadratic Equation by Factoring

Solve the quadratic equation by factoring: x 2 + 8 x + 15 = 0. x 2 + 8 x + 15 = 0.

Find two numbers whose product equals 15 15 and whose sum equals 8. 8. List the factors of 15. 15.

The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve.

The solutions are −3 −3 and −5. −5.

Solve the quadratic equation by factoring: x 2 − 4 x − 21 = 0. x 2 − 4 x − 21 = 0.

Using the Zero-Product Property to Solve a Quadratic Equation Written as the Difference of Squares

Solve the difference of squares equation using the zero-product property: x 2 − 9 = 0. x 2 − 9 = 0.

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.

The solutions are 3 3 and −3. −3.

Solve by factoring: x 2 − 25 = 0. x 2 − 25 = 0.

Solving a Quadratic Equation by Factoring when the Leading Coefficient is not 1

When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:

• With the quadratic in standard form, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , multiply a ⋅ c . a ⋅ c .
• Find two numbers whose product equals a c a c and whose sum equals b . b .
• Rewrite the equation replacing the b x b x term with two terms using the numbers found in step 2 as coefficients of x.
• Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
• Factor out the expression in parentheses.
• Set the expressions equal to zero and solve for the variable.

Solving a Quadratic Equation Using Grouping

Use grouping to factor and solve the quadratic equation: 4 x 2 + 15 x + 9 = 0. 4 x 2 + 15 x + 9 = 0.

First, multiply a c : 4 ( 9 ) = 36. a c : 4 ( 9 ) = 36. Then list the factors of 36. 36.

The only pair of factors that sums to 15 15 is 3 + 12. 3 + 12. Rewrite the equation replacing the b term, 15 x , 15 x , with two terms using 3 and 12 as coefficients of x . Factor the first two terms, and then factor the last two terms.

Solve using the zero-product property.

The solutions are − 3 4 , − 3 4 , and −3. −3. See Figure 3 .

Solve using factoring by grouping: 12 x 2 + 11 x + 2 = 0. 12 x 2 + 11 x + 2 = 0.

Solving a Polynomial of Higher Degree by Factoring

Solve the equation by factoring: −3 x 3 − 5 x 2 − 2 x = 0. −3 x 3 − 5 x 2 − 2 x = 0.

This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out − x − x from all of the terms and then proceed with grouping.

Use grouping on the expression in parentheses.

Now, we use the zero-product property. Notice that we have three factors.

The solutions are 0 , 0 , − 2 3 , − 2 3 , and −1. −1.

Solve by factoring: x 3 + 11 x 2 + 10 x = 0. x 3 + 11 x 2 + 10 x = 0.

Using the Square Root Property

When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property , in which we isolate the x 2 x 2 term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the x 2 x 2 term so that the square root property can be used.

The Square Root Property

With the x 2 x 2 term isolated, the square root property states that:

where k is a nonzero real number.

Given a quadratic equation with an x 2 x 2 term but no x x term, use the square root property to solve it.

• Isolate the x 2 x 2 term on one side of the equal sign.
• Take the square root of both sides of the equation, putting a ± ± sign before the expression on the side opposite the squared term.
• Simplify the numbers on the side with the ± ± sign.

Solving a Simple Quadratic Equation Using the Square Root Property

Solve the quadratic using the square root property: x 2 = 8. x 2 = 8.

Take the square root of both sides, and then simplify the radical. Remember to use a ± ± sign before the radical symbol.

The solutions are 2 2 , 2 2 , −2 2 . −2 2 .

Solving a Quadratic Equation Using the Square Root Property

Solve the quadratic equation: 4 x 2 + 1 = 7. 4 x 2 + 1 = 7.

First, isolate the x 2 x 2 term. Then take the square root of both sides.

The solutions are 6 2 , 6 2 , and − 6 2 . − 6 2 .

Solve the quadratic equation using the square root property: 3 ( x − 4 ) 2 = 15. 3 ( x − 4 ) 2 = 15.

• Completing the Square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square . Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a , must equal 1. If it does not, then divide the entire equation by a . Then, we can use the following procedures to solve a quadratic equation by completing the square.

We will use the example x 2 + 4 x + 1 = 0 x 2 + 4 x + 1 = 0 to illustrate each step.

Given a quadratic equation that cannot be factored, and with a = 1 , a = 1 , first add or subtract the constant term to the right side of the equal sign.

Multiply the b term by 1 2 1 2 and square it.

Add ( 1 2 b ) 2 ( 1 2 b ) 2 to both sides of the equal sign and simplify the right side. We have

The left side of the equation can now be factored as a perfect square.

Use the square root property and solve.

The solutions are −2 + 3 , −2 + 3 , and −2 − 3 . −2 − 3 .

Solving a Quadratic by Completing the Square

Solve the quadratic equation by completing the square: x 2 − 3 x − 5 = 0. x 2 − 3 x − 5 = 0.

First, move the constant term to the right side of the equal sign.

Then, take 1 2 1 2 of the b term and square it.

Add the result to both sides of the equal sign.

Factor the left side as a perfect square and simplify the right side.

The solutions are 3 + 29 2 3 + 29 2 and 3 - 29 2 3 - 29 2 .

Solve by completing the square: x 2 − 6 x = 13. x 2 − 6 x = 13.

Using the Quadratic Formula

The fourth method of solving a quadratic equation is by using the quadratic formula , a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

We can derive the quadratic formula by completing the square . We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by −1 −1 and obtain a positive a . Given a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , a ≠ 0 , a ≠ 0 , we will complete the square as follows:

First, move the constant term to the right side of the equal sign:

As we want the leading coefficient to equal 1, divide through by a :

Then, find 1 2 1 2 of the middle term, and add ( 1 2 b a ) 2 = b 2 4 a 2 ( 1 2 b a ) 2 = b 2 4 a 2 to both sides of the equal sign:

Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:

Now, use the square root property, which gives

Finally, add − b 2 a − b 2 a to both sides of the equation and combine the terms on the right side. Thus,

The Quadratic Formula

Written in standard form, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , any quadratic equation can be solved using the quadratic formula :

where a , b , and c are real numbers and a ≠ 0. a ≠ 0.

Given a quadratic equation, solve it using the quadratic formula

• Make sure the equation is in standard form: a x 2 + b x + c = 0. a x 2 + b x + c = 0.
• Make note of the values of the coefficients and constant term, a , b , a , b , and c . c .
• Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
• Calculate and solve.

Solve the Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation: x 2 + 5 x + 1 = 0. x 2 + 5 x + 1 = 0.

Identify the coefficients: a = 1 , b = 5 , c = 1. a = 1 , b = 5 , c = 1. Then use the quadratic formula.

Solving a Quadratic Equation with the Quadratic Formula

Use the quadratic formula to solve x 2 + x + 2 = 0. x 2 + x + 2 = 0.

First, we identify the coefficients: a = 1 , b = 1 , a = 1 , b = 1 , and c = 2. c = 2.

Substitute these values into the quadratic formula.

The solutions to the equation are − 1 + i 7 2 − 1 + i 7 2 and − 1 − i 7 2 − 1 − i 7 2

Solve the quadratic equation using the quadratic formula: 9 x 2 + 3 x − 2 = 0. 9 x 2 + 3 x − 2 = 0.

The Discriminant

The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant , or the expression under the radical, b 2 − 4 a c . b 2 − 4 a c . The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table 1 relates the value of the discriminant to the solutions of a quadratic equation.

For a x 2 + b x + c = 0 a x 2 + b x + c = 0 , where a a , b b , and c c are real numbers, the discriminant is the expression under the radical in the quadratic formula: b 2 − 4 a c . b 2 − 4 a c . It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.

Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation

Use the discriminant to find the nature of the solutions to the following quadratic equations:

• ⓐ x 2 + 4 x + 4 = 0 x 2 + 4 x + 4 = 0
• ⓑ 8 x 2 + 14 x + 3 = 0 8 x 2 + 14 x + 3 = 0
• ⓒ 3 x 2 − 5 x − 2 = 0 3 x 2 − 5 x − 2 = 0
• ⓓ 3 x 2 − 10 x + 15 = 0 3 x 2 − 10 x + 15 = 0

Calculate the discriminant b 2 − 4 a c b 2 − 4 a c for each equation and state the expected type of solutions.

x 2 + 4 x + 4 = 0 x 2 + 4 x + 4 = 0

b 2 − 4 a c = ( 4 ) 2 − 4 ( 1 ) ( 4 ) = 0. b 2 − 4 a c = ( 4 ) 2 − 4 ( 1 ) ( 4 ) = 0. There will be one rational double solution.

8 x 2 + 14 x + 3 = 0 8 x 2 + 14 x + 3 = 0

b 2 − 4 a c = ( 14 ) 2 − 4 ( 8 ) ( 3 ) = 100. b 2 − 4 a c = ( 14 ) 2 − 4 ( 8 ) ( 3 ) = 100. As 100 100 is a perfect square, there will be two rational solutions.

3 x 2 − 5 x − 2 = 0 3 x 2 − 5 x − 2 = 0

b 2 − 4 a c = ( −5 ) 2 − 4 ( 3 ) ( −2 ) = 49. b 2 − 4 a c = ( −5 ) 2 − 4 ( 3 ) ( −2 ) = 49. As 49 49 is a perfect square, there will be two rational solutions.

3 x 2 −10 x + 15 = 0 3 x 2 −10 x + 15 = 0

b 2 − 4 a c = ( −10 ) 2 − 4 ( 3 ) ( 15 ) = −80. b 2 − 4 a c = ( −10 ) 2 − 4 ( 3 ) ( 15 ) = −80. There will be two complex solutions.

Using the Pythagorean Theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem . It is based on a right triangle, and states the relationship among the lengths of the sides as a 2 + b 2 = c 2 , a 2 + b 2 = c 2 , where a a and b b refer to the legs of a right triangle adjacent to the 90° 90° angle, and c c refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.

We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.

The Pythagorean Theorem is given as

where a a and b b refer to the legs of a right triangle adjacent to the 90 ∘ 90 ∘ angle, and c c refers to the hypotenuse, as shown in Figure 4 .

Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle in Figure 5 .

As we have measurements for side b and the hypotenuse, the missing side is a.

Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.

Access these online resources for additional instruction and practice with quadratic equations.

• The Zero-Product Property
• Quadratic Formula with Two Rational Solutions
• Length of a leg of a right triangle

2.5 Section Exercises

How do we recognize when an equation is quadratic?

When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why when solving a quadratic equation in the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 we may graph the equation y = a x 2 + b x + c y = a x 2 + b x + c and have no zeroes ( x -intercepts).

When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?

In the quadratic formula, what is the name of the expression under the radical sign b 2 − 4 a c , b 2 − 4 a c , and how does it determine the number of and nature of our solutions?

Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.

For the following exercises, solve the quadratic equation by factoring.

x 2 + 4 x − 21 = 0 x 2 + 4 x − 21 = 0

x 2 − 9 x + 18 = 0 x 2 − 9 x + 18 = 0

2 x 2 + 9 x − 5 = 0 2 x 2 + 9 x − 5 = 0

6 x 2 + 17 x + 5 = 0 6 x 2 + 17 x + 5 = 0

4 x 2 − 12 x + 8 = 0 4 x 2 − 12 x + 8 = 0

3 x 2 − 75 = 0 3 x 2 − 75 = 0

8 x 2 + 6 x − 9 = 0 8 x 2 + 6 x − 9 = 0

4 x 2 = 9 4 x 2 = 9

2 x 2 + 14 x = 36 2 x 2 + 14 x = 36

5 x 2 = 5 x + 30 5 x 2 = 5 x + 30

4 x 2 = 5 x 4 x 2 = 5 x

7 x 2 + 3 x = 0 7 x 2 + 3 x = 0

x 3 − 9 x = 2 x 3 − 9 x = 2

For the following exercises, solve the quadratic equation by using the square root property.

x 2 = 36 x 2 = 36

x 2 = 49 x 2 = 49

( x − 1 ) 2 = 25 ( x − 1 ) 2 = 25

( x − 3 ) 2 = 7 ( x − 3 ) 2 = 7

( 2 x + 1 ) 2 = 9 ( 2 x + 1 ) 2 = 9

( x − 5 ) 2 = 4 ( x − 5 ) 2 = 4

For the following exercises, solve the quadratic equation by completing the square. Show each step.

x 2 − 9 x − 22 = 0 x 2 − 9 x − 22 = 0

2 x 2 − 8 x − 5 = 0 2 x 2 − 8 x − 5 = 0

x 2 − 6 x = 13 x 2 − 6 x = 13

x 2 + 2 3 x − 1 3 = 0 x 2 + 2 3 x − 1 3 = 0

2 + z = 6 z 2 2 + z = 6 z 2

6 p 2 + 7 p − 20 = 0 6 p 2 + 7 p − 20 = 0

2 x 2 − 3 x − 1 = 0 2 x 2 − 3 x − 1 = 0

For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.

2 x 2 − 6 x + 7 = 0 2 x 2 − 6 x + 7 = 0

x 2 + 4 x + 7 = 0 x 2 + 4 x + 7 = 0

3 x 2 + 5 x − 8 = 0 3 x 2 + 5 x − 8 = 0

9 x 2 − 30 x + 25 = 0 9 x 2 − 30 x + 25 = 0

2 x 2 − 3 x − 7 = 0 2 x 2 − 3 x − 7 = 0

6 x 2 − x − 2 = 0 6 x 2 − x − 2 = 0

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution .

2 x 2 + 5 x + 3 = 0 2 x 2 + 5 x + 3 = 0

x 2 + x = 4 x 2 + x = 4

3 x 2 − 5 x + 1 = 0 3 x 2 − 5 x + 1 = 0

x 2 + 4 x + 2 = 0 x 2 + 4 x + 2 = 0

4 + 1 x − 1 x 2 = 0 4 + 1 x − 1 x 2 = 0

For the following exercises, enter the expressions into your graphing utility and find the zeroes to the equation (the x -intercepts) by using 2 nd CALC 2:zero . Recall finding zeroes will ask left bound (move your cursor to the left of the zero,enter), then right bound (move your cursor to the right of the zero,enter), then guess (move your cursor between the bounds near the zero, enter). Round your answers to the nearest thousandth.

Y 1 = 4 x 2 + 3 x − 2 Y 1 = 4 x 2 + 3 x − 2

Y 1 = −3 x 2 + 8 x − 1 Y 1 = −3 x 2 + 8 x − 1

Y 1 = 0.5 x 2 + x − 7 Y 1 = 0.5 x 2 + x − 7

To solve the quadratic equation x 2 + 5 x − 7 = 4 , x 2 + 5 x − 7 = 4 , we can graph these two equations

Y 1 = x 2 + 5 x − 7 Y 2 = 4 Y 1 = x 2 + 5 x − 7 Y 2 = 4

and find the points of intersection. Recall 2 nd CALC 5:intersection. Do this and find the solutions to the nearest tenth.

To solve the quadratic equation 0.3 x 2 + 2 x − 4 = 2 , 0.3 x 2 + 2 x − 4 = 2 , we can graph these two equations

Y 1 = 0.3 x 2 + 2 x − 4 Y 2 = 2 Y 1 = 0.3 x 2 + 2 x − 4 Y 2 = 2

Beginning with the general form of a quadratic equation, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , solve for x by using the completing the square method, thus deriving the quadratic formula.

Show that the sum of the two solutions to the quadratic equation is − b a − b a .

A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 119 ft. 2 . Solve the quadratic equation to find the length and width.

Abercrombie and Fitch stock had a price given as P = 0.2 t 2 − 5.6 t + 50.2 , P = 0.2 t 2 − 5.6 t + 50.2 , where t t is the time in months from 1999 to 2001. ( t = 1 t = 1 is January 1999). Find the two months in which the price of the stock was $30.

Suppose that an equation is given p = −2 x 2 + 280 x − 1000 , p = −2 x 2 + 280 x − 1000 , where x x represents the number of items sold at an auction and p p is the profit made by the business that ran the auction. How many items sold would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2 nd CALC maximum. To obtain a good window for the curve, set x x [0,200] and y y [0,10000].

Real-World Applications

A formula for the normal systolic blood pressure for a man age A , A , measured in mmHg, is given as P = 0.006 A 2 − 0.02 A + 120. P = 0.006 A 2 − 0.02 A + 120. Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.

The cost function for a certain company is C = 60 x + 300 C = 60 x + 300 and the revenue is given by R = 100 x − 0.5 x 2 . R = 100 x − 0.5 x 2 . Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $300.

A falling object travels a distance given by the formula d = 5 t + 16 t 2 d = 5 t + 16 t 2 ft, where t t is measured in seconds. How long will it take for the object to travel 74 ft?

A vacant lot is being converted into a community garden. The garden and the walkway around its perimeter have an area of 378 ft 2 . Find the width of the walkway if the garden is 12 ft. wide by 15 ft. long.

An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P P , who contracted the flu t t days after it broke out is given by the model P = − t 2 + 13 t + 130 , P = − t 2 + 13 t + 130 , where 1 ≤ t ≤ 6. 1 ≤ t ≤ 6. Find the day that 160 students had the flu. Recall that the restriction on t t is at most 6.

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Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites

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• Book title: College Algebra 2e
• Publication date: Dec 21, 2021
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• Section URL: https://openstax.org/books/college-algebra-2e/pages/2-5-quadratic-equations

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• Solve Quadratic Equations by Factoring
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• Quadratic Formula Worksheet (real solutions)
• Quadratic Formula Worksheet (complex solutions)
• Quadratic Formula Worksheet (both real and complex solutions)
• Discriminant Worksheet
• Sum and Product of Roots
• Radical Equations Worksheet

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• RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations
• RS Aggarwal Solutions

Class 11 RS Aggarwal Chapter-5 Complex Numbers and Quadratic Equations Solutions - Free PDF Download

The best part about the Vedantu website is that it lets you download the Complex Number Class 11 RS Aggarwal solutions for free. All that you need is a device where you can download and refer to the solution on the go even when you have no internet. This makes it very handy. You can also print out a hard copy and refer to the pdf solutions when you wish to.

Complex Numbers Class 11 RS Aggarwal Solutions

The Complex Number Class 11 RS Aggarwal solutions talk about what complex numbers are. A complex number is nothing but a number that can be expressed in the form of p+iq. This is where both p and q are real numbers, and i is the solution of any equation. The numbers 8+2i and 2+3i are examples of complex numbers.

Generally, the complex number will be represented as z = p + iq

Here p is the real part that is denoted by Re z and q is the imaginary part, and it is denoted by Im z

A complex number will be purely real when its imaginary part is equal to 0.

The complex number also follows the concept of equality,

Two complex numbers z 1 = a 1 + ib 1 and z 2 = a 2 + ib 2 are equal, if a 1 = a 2 and b 1 = b 2 i.e., Re (z 1 ) = Re (z 2 ) and Im (z 1 ) = Im (z 2 ).

The complex number can also be added. Suppose there are two numbers like Let z 1 = m + ni and z 2 = o + ip. These are two kinds of complex numbers. You can add both zi and z 2 . So basically z 1 + z 2 = z = (m + o) + (n + p)I and z that is obtained is the complex number that is obtained.

The sum of complex numbers will always be a complex number as per the closure law.

For complex numbers z 1 and z 2 : z 2 + z 1 = z 1 + z 2 as per the commutative law)

 For complex numbers z 1 , z 2 , z 3 : (z 1 + z 2 ) + z 3 = z 1 + (z 2 + z 3 ) . This is the associative law.

For every complex number z, z + 0 = z. This is the additive identity

To every complex number z = p + qi, then we have the complex number -z = -p + i(-q), called the negative or additive inverse of z.

z +(– z )=0

The complex numbers can also be subtracted from each other. So if there are two complex numbers z 1 = m + ni and z 2 = o + ip then you can calculate z 1 – z 2 = z 1 + (-z 2 ).

Complex numbers can also be multiplied. If z 1 = m + ni and z 2 = o + ip are two complex numbers then z 1 × z 2 = (mo – np) + i(no + pm)

For the complex number multiplication, you need to remember these.

For complex numbers z 1 and z 2 , z 1 × z 2 = z 2 × z 1 (commutative law).

For complex numbers z 1 , z 2 , z 3 , (z 1 × z 2 ) × z 3 = z 1 × (z 2 × z 3 )

associative law

associative law.

Here are the properties of complex number multiplication

Commutative z 1 z 2 = z 2 z 1

(ii) Associative (z 1 z 2 ) z 3 = z 1 (z 2 z 3 )

(iii) Multiplicative Identity z • 1 = z = 1 • z

Here, 1 is the multiplicative identity of an element z.

(iv) Multiplicative Inverse Every non-zero complex number z there exists a complex number z 1 such that z.z 1     = 1 = z 1 • z

(v)  Distributive Law

(a)  z 1 (z 2 + z 3 ) = z 1 z 2 + z 1 z 3   (left distribution)

(b) (z 2 + z 3 )z 1 = z 2 z 1 + z 3 z 1 (right distribution)

Also when z 1 = m + in and z 2 = o + ip. Then,

z 1 + z 2 = (m + o) + i (n + p)

z 1 z 2 = (mo – np) + i(mp + on)

The conjugate of any complex number z = m + in, denoted by z¯¯¯, is given by z = m – in.

The modulus and the conjugate of complex numbers state that when z = m + in is a complex number then the modulus of z, denoted by |z| = m 2 −n 2 −−−−−−−√ and the conjugate of z, denoted by z¯¯¯ is the complex number m – ni.

Preparation Tips for RS Aggarwal Class 11 Math Chapter 5

Practicing these solutions on Complex Numbers is important because it forms the key part in building basics to perform well in mathematics in higher classes.

Going through these RS Aggarwal Class 11 Solutions Complex Numbers solutions will get you prepared to do complex problems on complex numbers with ease either in your school or competitive examinations

The topic is covered with detailed solutions on this website that builds on your confidence.

RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations are a must for those students who are having trouble with the chapter concepts. Quadratic Equation is not only an important concept from the CBSE exams point of view but is also an important part of a lot of exams such as NEET, JEE, IPU-CET, etc. Chapter 5 of RS Aggarwal Solutions will provide you with a detailed idea regarding the topics involved such as the arithmetic operations on complex numbers, finding roots of quadratic equations, and determining their nature. RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations provides a deep analysis of the chapter along with various handy examples.  

Benefits Of Solving Exercises Through Rs Aggarwal Class 11 Solutions Chapter-5 Complex Numbers And Quadratic Equations

If you want to solve the questions from the exercises of Chapter 5 Complex Numbers and Quadratic Equations then it is highly recommended to check out RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations for all the solutions that you wish for.

It is a great reference book when it comes to preparing for exams such as JEE or even NEET along with your class 11 tests. Vedantu is always here to help you out if there are any problems while solving the questions.

FAQs on RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations

1. What are some of the important points covered in RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations that should be kept in mind before the Class 11 exam?

The important points that are covered in the RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations are as follows:

The general form of a complex number: z=p+ iq, where p is called the real part, denoted by Re z, q is known as the imaginary part and is denoted by Im z.

Two complex numbers are said to be real if they have a sum and a product that is considered to be real.

Any equation that consists of a variable of the highest degree 2 is called the quadratic equation.

An equation of degree n has a total of n roots, either real or imaginary.

An odd degree tends to have at least one real root whose sign is opposite to that of its last term, also called the constant term, provided that the coefficient of the highest degree term is positive.

2. Is just copying from RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations an option when it comes to homework?

While copying from the RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations is an option to finish off your pending homework, it won't help you with studying and understanding concepts. To get good marks in the exam, it is important that while doing your homework you understand each concept that has been mentioned in the book and then continue to solve the problems. In such a way you will also be able to solve the question for the Class 11 exam on your own.

3. Will studying from RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations relieve me from stress?

Stress is created when a person tries to do a tremendous amount of activities in a small amount of time. Doing so will not only deteriorate your health but you will also end up losing your focus. Hence if you are studying for the exams then make sure you refer to the RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations and understand the concepts well beforehand so that there is no stress when you will be giving your exams. The easy language and thorough exams truly help students feel confident about the subject

4. What makes the RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations a special book to refer to?

The RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations is a special book to refer to due to the following conditions:

It has everything you need to know about before the exams start or even when you are just practicing

It helps you with solving some competitive problems from other exams like JEE which also helps you challenge yourself.

You get a strong foundation when you understand concepts right from Class 11 as it helps you learn better in Class 12.  

5. Are there any previous years’ papers included in the RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations that will help me brush up before the exam?

Yes, if a student wishes to understand how each part of the syllabus carries the marks weightage then he or she can take a reference of the previous years’ papers provided in the RS Aggarwal Class 11 Solutions Chapter-5 Complex Numbers and Quadratic Equations. Taking a look and solving these papers will help students understand the concepts and apply these to any type of questions they might be asked during the exam.  Practicing the Vedantu CBSE Class 11 Sample paper with Solutions will also help you get an idea of the exam-giving pattern.

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• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
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• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

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• High School Math Solutions – Quadratic Equations Calculator, Part 3 On the last post we covered completing the square (see link). It is pretty strait forward if you follow all the...

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