# How do you solve (p-4)^2=16?

Aug 31, 2016

$p = 8 \mathmr{and} p = 0$

#### Explanation:

Although this is a quadratic, it is a special case in that there is no term in $x$.

We can simply find the square root of each side,

$\sqrt{{\left(p - 4\right)}^{2}} = \pm \sqrt{16}$

$p - 4 = \pm 4$

$p - 4 = + 4 \rightarrow p = 8$

$p - 4 = - 4 \rightarrow p = 0$

Aug 31, 2016

P=0 or 8

#### Explanation:

${\left(p - 4\right)}^{2} = 16$
Take square root of both sides
$\left(p - 4\right) = \pm 4$
So p-4=-4 which gives p=0

Or
p-4=4 which gives p=8