# How do you solve (x-2)^2-25=0?

Aug 24, 2016

$x = 7$

$x = - 3$

#### Explanation:

${\left(x - 2\right)}^{2} - 25 = 0$

or

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

or

$x - 2 = \sqrt{25}$

or

$x - 2 = \pm 5$

or

$x = 2 \pm 5$

or

$x = 2 + 5$

or

$x = 7$---------------------------------Ans $1$

or

$x = 2 - 5$

or

$x = - 3$-----------------------------Ans $2$

Aug 24, 2016

$x = 7 , x = - 3$

#### Explanation:

First step is to isolate the ${\left(x - 2\right)}^{2}$ term

$\Rightarrow {\left(x - 2\right)}^{2} - \cancel{25} + \cancel{25} = 0 + 25$

$\Rightarrow {\left(x - 2\right)}^{2} = 25$

now take the $\textcolor{b l u e}{\text{square root of both sides}}$

$\sqrt{{\left(x - 2\right)}^{2}} = \pm \sqrt{25}$

$\Rightarrow x - 2 = \pm 5$

solve : $x - 2 = 5 \Rightarrow x = 5 + 2 = 7$

solve : $x - 2 = - 5 \Rightarrow x = - 5 + 2 = - 3$

Aug 24, 2016

$x = 7$ or $x = - 3$

#### Explanation:

We can use identity $\left({a}^{2} - {b}^{2}\right) = \left(a - b\right) \left(a + b\right)$. As such

${\left(x - 2\right)}^{2} - 25 = 0$

$\Leftrightarrow {\left(x - 2\right)}^{2} - {5}^{2} = 0$

or $\left(x - 2 - 5\right) \left(x - 2 + 5\right) = 0$

or $\left(x - 7\right) \left(x + 3\right) = 0$

i.e. $x = 7$ or $x = - 3$