# How do you solve 25x^2-64?

Jun 28, 2015

This is a difference of squares:

$25 {x}^{2} - 64 = {\left(5 x\right)}^{2} - {8}^{2} = \left(5 x - 8\right) \left(5 x + 8\right)$

So the roots of $25 {x}^{2} - 64 = 0$ are $\frac{8}{5}$ and $- \frac{8}{5}$

#### Explanation:

Any difference of squares is of the form ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

In our case $a = 5 x$ and $b = 8$, giving:

$25 {x}^{2} - 64 = {\left(5 x\right)}^{2} - {8}^{2}$

$= {a}^{2} - {b}^{2}$

$= \left(a - b\right) \left(a + b\right)$

$= \left(5 x - 8\right) \left(5 x + 8\right)$