# Unit 11: Quadratic Equations and Functions

Check this section number; it may not be 11.1. Also it's not dynamic in any way (and so the button's disabled).

Let's look at solving a quadratic equation with a leading coefficient: $ax^2+bx+c=0$.

Example

Solve $6x^2-x-2=0$.

Identify $a$, $b$, and $c$.

$a=6$, $b=-1$, $c=-2$.

Since $a \ne 1$, we need to use a longer process.

Multiply $a$ and $c$.

$ac=\left(6\right)\left(-2\right)=-12$

Find factors of $ac$ which add to $b$.

We need to find a pair of numbers which add to $b=-1$ and multiply to $ac=-12$.

Because $ac$ is negative, we need one positive and one negative factor:

The pair of numbers we've found is $-4$ and $3$.

Rewrite the middle term of our equation using our newly-discovered factors.

$6x^2-4x+3x-2=0$.

Factor the two pairs of terms.

$\underbrace{6x^2-4x}_{2x\left(3x-2\right)} + \underbrace{3x-2}_{1\left(3x-2\right)} =0$.

Rewrite using grouping.

...

Solve each part

\underbrace{\left(2x+1\right)}_{ \begin{aligned}2x+1&=0\\2x&=-1\\x&=-\frac{1}{2}\end{aligned} }\underbrace{\left(3x-2\right)}_{ \begin{aligned}3x-2&=0\\3x&=2\\x&=\frac{2}{3}\end{aligned} }=0.

We have the two answers: $x=-\frac{1}{2}$ and $x=\frac{2}{3}$.