• Anatomy & Physiology
  • Astrophysics
  • Earth Science
  • Environmental Science
  • Organic Chemistry
  • Precalculus
  • Trigonometry
  • English Grammar
  • U.S. History
  • World History

... and beyond

  • Socratic Meta
  • Featured Answers

Search icon

  • Formulas for Problem Solving

Key Questions

A few examples...

Explanation:

I will assume that you mean things like common identities and the quadratic formula . Here are just a few:

Difference of squares identity

#a^2-b^2 = (a-b)(a+b)#

Deceptively simple, but massively useful.

For example:

#a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2# #color(white)(a^4+b^4) = (a^2+b^2)^2 - (sqrt(2)ab)^2# #color(white)(a^4+b^4) = ((a^2+b^2) - sqrt(2)ab)((a^2+b^2) +sqrt(2)ab)# #color(white)(a^4+b^4) = (a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)#

Difference of cubes identity

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

Sum of cubes identity

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Quadratic formula

Very useful to know, better if you know how to derive it:

The zeros of #ax^2+bx+c# are given by:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

Pythagoras theorem

If a right angled triangle has legs of length #a, b# and hypotenuse of length #c# then:

#c^2 = a^2+b^2#

This is also very useful in trigonometric form. If we have an angle #theta# in a right-angled triangle, then we call the side nearest #theta# , the #"adjacent"# side, the side opposite it the #"opposite"# side and the hypotenuse the #"hypotenuse"# .

#"hypotenuse"^2 = "adjacent"^2 + "opposite"^2#

Dividing both sides by #"hypotenuse"^2# , we get:

#1 = ("adjacent"/"hypotenuse")^2 + ("opposite"/"hypotenuse")^2#
#1 = cos^2 theta + sin^2 theta#

Then dividing both sides by #cos^2 theta# we find:

#sec^2 theta = 1 + tan^2 theta#

Binomial theorem

#(a+b)^n = sum_(k=0)^n ((n), (k)) a^(n-k) b^k#

where #((n), (k)) = (n!)/((n-k)! k!)#

#(x+1)^4 = x^4+4x^3+6x^2+4x+1#

formula of problem solving

Using Pythagoras' Theorem, we know #A^2 + B^2 = C^2#

#A^2 = 9# , #B^2 = 16 implies A^2 + B^2 = 9 + 16 = 25#

SInce #C^2 = A^2 + B^2 = 25# , you know that

#C = sqrt(C^2) = sqrt(25) = 5#

You simply find out what value goes with what part of the formula, and do this for each, and then work out the formula as normal.

formula of problem solving

IMAGES

  1. PPT

    formula of problem solving

  2. PPT

    formula of problem solving

  3. Formulas and Problem Solving

    formula of problem solving

  4. Problem Solving: Using Formulas

    formula of problem solving

  5. Introduction to Problem Solving Skills

    formula of problem solving

  6. The 5 Steps of Problem Solving

    formula of problem solving

VIDEO

  1. Electricity| CBSE 10| Part 10| Physics| Tamil| Sharmila Ma'am

  2. Cauchy Integral Formula(Problem Solving)L4

  3. Pure Maths EQ

  4. 37. Gaussian Quadrature(Formula + Problem Solving)||BSc Honours-2nd year||Bangla

  5. Solving a Formula Problem in Algebra

  6. 7.6 Surface Area and Volume of Cylinders