# How do you solve (x-5)^2=4?

Mar 7, 2017

#### Answer:

$\text{the answer : x=3 or x=7}$

#### Explanation:

${\left(x - 5\right)}^{2} = 4$

$\sqrt{{\left(x - 5\right)}^{2}} = \sqrt{4}$

$\pm \left(x - 5\right) = \pm 2$

$\text{if "+(x-5)=+2" then "x-5=2" , } x = 7$

$\text{if "+(x-5)=-2" then "x-5=-2" , } x = 3$

$\text{if "-(x-5)=+2" then "-x+5=2" , } x = 3$

$\text{if "-(x-5)=-2" then "-x+5=-2" , } x = 7$

$\text{thus ;}$

$\text{x=3 or x=7}$

Mar 7, 2017

#### Answer:

$x = 7$ & $x = 3$

#### Explanation:

• Method 1-

${\left(x - 5\right)}^{2} = 4$

$\Rightarrow x - 5 = \pm 2$ ----(taking square root both sides)

$\Rightarrow x - 5 = 2$ & $\Rightarrow x - 5 = - 2$

$\Rightarrow x = 7$ & $\Rightarrow x = 3$

• Method 2-

${\left(x - 5\right)}^{2} = 4$

$\Rightarrow {x}^{2} - 10 x + 25 = 4$

$\Rightarrow {x}^{2} - 10 x + 21 = 0$

$\Rightarrow {x}^{2} - 7 x - 3 x + 21 = 0$

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

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

$\Rightarrow x = 7$ & $\Rightarrow x = 3$

• Method 3-

Graphical Approach

Plot $y = {\left(x - 5\right)}^{2}$ and $y = 4$ on graph paper. Their intersection points are its solutions.

$y = {\left(x - 5\right)}^{2}$ $\rightarrow$ Parabola
$y = 4$$\rightarrow$ Straight line parallel to x-axis.

Mar 7, 2017

#### Answer:

$x = 3 \text{ or } x = 7$

#### Explanation:

To 'undo' the square on the left side.

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

$\sqrt{{\left(x - 5\right)}^{2}} = \pm \sqrt{4}$

$\Rightarrow x - 5 = \pm 2 \leftarrow \textcolor{red}{\text{ there are 2 solutions}}$

• x-5=+2

$\Rightarrow x = 2 + 5$

$\Rightarrow x = 7 \to \left(\textcolor{red}{1}\right)$

• x-5=-2

$\Rightarrow x = - 2 + 5$

$\Rightarrow x = 3 \to \left(\textcolor{red}{2}\right)$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

$x = 7 \to {\left(7 - 5\right)}^{2} = 4 = \text{ right side}$

$x = 3 \to {\left(3 - 5\right)}^{2} = 4 = \text{ right side}$

$\Rightarrow x = 3 \text{ or "x=7" are the solutions}$