# How do you factor and solve 2x^2 - 3= 125?

Apr 1, 2018

$x = \pm 8$

#### Explanation:

$2 {x}^{2} - 3 = 125$

Subtract 125 on both sides

$2 {x}^{2} - 128$=0

Divide both sides by 2

${x}^{2} - 64 = 0$

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

${x}^{2} - 64 = \left(x + 8\right) \left(x - 8\right)$

So $\left(x + 8\right) \left(x - 8\right) = 0$

$x = \pm 8$

Apr 1, 2018

$2 {x}^{2} - 3 = 125$ can be factored to:

$2 \left(x - 8\right) \left(x + 8\right) = 0$, and has the solution:

$\textcolor{red}{\left\mid x \right\mid = 8}$

#### Explanation:

Move all the terms to one side of the equation

$2 {x}^{2} - 3 = 125$

$2 {x}^{2} - 3 - \textcolor{red}{125} = \cancel{125} - \cancel{\textcolor{red}{125}}$

$2 {x}^{2} - 128 = 0$

Now take out a factor of 2

$\left(\textcolor{red}{2} \cdot {x}^{2}\right) - \left(\textcolor{red}{2} \cdot 64\right) = 0$

$\textcolor{red}{2} \left({x}^{2} - 64\right) = 0$

We now have a term in the parentheses that looks like

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

This is called a difference of squares

We can factor a difference of squares like this:

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

Let's apply this to our expression

$2 \left({x}^{2} - \textcolor{red}{64}\right) = 0$

$2 \left({x}^{2} - \textcolor{red}{{8}^{2}}\right) = 0$

$2 \left(x - 8\right) \left(x + 8\right) = 0$

This is the fully factored form.

By examining this equation, we can see that the solutions — the values of $x$ that make the equation true — are

$x = 8$

and

$x = - 8$

or simply

$\left\mid x \right\mid = 8$