What are commonly used formulas used in problem solving?

1 Answer
Feb 10, 2018

Answer:

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"#.

Then:

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

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

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

That is:

#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!)#

For example:

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