# How do you solve (x-3)^2 = 5?

Jun 27, 2018

See a solution process below:

#### Explanation:

First, put the equation in standard quadratic form:

${\left(x - 3\right)}^{2} = 5$

$\left(x - 3\right) \left(x - 3\right) = 5$

${x}^{2} - 3 x - 3 x + 9 = 5$

${x}^{2} - 6 x + 9 = 5$

${x}^{2} - 6 x + 9 - \textcolor{red}{5} = 5 - \textcolor{red}{5}$

${x}^{2} - 6 x + 4 = 0$

We can now use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 6}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{4}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 6} \pm \sqrt{{\textcolor{b l u e}{- 6}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{4}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{6 \pm \sqrt{36 - 16}}{2}$

$x = \frac{6 \pm \sqrt{20}}{2}$

$x = \frac{6 - \sqrt{4 \cdot 5}}{2}$; $x = \frac{6 + \sqrt{4 \cdot 5}}{2}$

$x = \frac{6 - \sqrt{4} \sqrt{5}}{2}$; $x = \frac{6 + \sqrt{4} \sqrt{5}}{2}$

$x = \frac{6 - 2 \sqrt{5}}{2}$; $x = \frac{6 + 2 \sqrt{5}}{2}$

$x = \frac{6}{2} - \frac{2 \sqrt{5}}{2}$; $x = \frac{6}{2} + \frac{2 \sqrt{5}}{2}$

$x = 3 - \sqrt{5}$; $x = 3 + \sqrt{5}$

Jun 27, 2018

$x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}$

#### Explanation:

Let's take the square root of both sides. This gives us

$x - 3 = \pm \sqrt{5}$

We can add $3$ to both sides to get

$x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}$

Hope this helps!