# 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

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Deceptively simple, but massively useful.

For example:

${a}^{4} + {b}^{4} = {\left({a}^{2} + {b}^{2}\right)}^{2} - 2 {a}^{2} {b}^{2}$

$\textcolor{w h i t e}{{a}^{4} + {b}^{4}} = {\left({a}^{2} + {b}^{2}\right)}^{2} - {\left(\sqrt{2} a b\right)}^{2}$

$\textcolor{w h i t e}{{a}^{4} + {b}^{4}} = \left(\left({a}^{2} + {b}^{2}\right) - \sqrt{2} a b\right) \left(\left({a}^{2} + {b}^{2}\right) + \sqrt{2} a b\right)$

$\textcolor{w h i t e}{{a}^{4} + {b}^{4}} = \left({a}^{2} - \sqrt{2} a b + {b}^{2}\right) \left({a}^{2} + \sqrt{2} a b + {b}^{2}\right)$

Difference of cubes identity

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Sum of cubes identity

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

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

The zeros of $a {x}^{2} + b x + c$ are given by:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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 $\text{adjacent}$ side, the side opposite it the $\text{opposite}$ side and the hypotenuse the $\text{hypotenuse}$.

Then:

${\text{hypotenuse"^2 = "adjacent"^2 + "opposite}}^{2}$

Dividing both sides by ${\text{hypotenuse}}^{2}$, we get:

$1 = {\left(\text{adjacent"/"hypotenuse")^2 + ("opposite"/"hypotenuse}\right)}^{2}$

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

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

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

For example:

${\left(x + 1\right)}^{4} = {x}^{4} + 4 {x}^{3} + 6 {x}^{2} + 4 x + 1$

• Hmmm, there is no rule when you have a word math problem and also there is no specific forumula that you use when you need to solve word problem. It is all about logic. If you want, you can post question (specific word problem) and we will anwer it and explain how to solve it.

#### Explanation:

The way you interpret the problem is the start, let me give you an example:

Triangle "1" Has 3 sides, $A$, $B$, and $C$. $A$ Is equal to $3$, $B$ is equal to $4$. Find $C$.

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.