# How do you solve 4x^2 - 16 = 0?

Apr 23, 2018

$x = \pm 2$

#### Explanation:

1. Add $16$ to both sides to get $4 {x}^{2} = 16$
2. Divide both sides by $4$ to get ${x}^{2} = 4$
3. Find the square root of both sides to get $x = \pm 2$
Apr 24, 2018

$x = \pm 2$

#### Explanation:

$4 {x}^{2} - 16$ fits the difference of squares property

${a}^{2} - {b}^{2}$, which can be factored as $\left(a + b\right) \left(a - b\right)$

Here, we can take the square root of $4 {x}^{2}$, which is $2 x$ and we can take the square root of $16$ which is $\pm 4$.

Thus, $a = 2 x$ and $b = \pm 4$, so we have

$\left(2 x + 4\right) \left(2 x - 4\right) = 0$

We can factor a $2$ out of both expressions to get

$2 \left(x + 2\right) \left(x - 2\right)$

Setting the factors equal to zero, we get

$x = \pm 2$

Hope this helps!