# How do you solve \sqrt{x^2-5x}-6=0?

May 20, 2018

$x = - 4 \mathmr{and} x = 9$

#### Explanation:

$\sqrt{{x}^{2} - 5 x} - 6 = 0$

$\sqrt{{x}^{2} - 5 x} = 6$

${\left(\sqrt{{x}^{2} - 5 x}\right)}^{2} = {6}^{2}$

${x}^{2} - 5 x = 36$

${x}^{2} - 5 x - 36 = 0$. factor.

$\left(x + 4\right) \left(x - 9\right)$

$x = - 4 \mathmr{and} x = 9$

May 20, 2018

$x = - 4 \text{ or } x = 9$

#### Explanation:

$\text{isolate "sqrt(x^2-5x)" by adding 6 to both sides}$

$\Rightarrow \sqrt{{x}^{2} - 5 x} = 6$

$\textcolor{b l u e}{\text{square both sides}}$

(sqrt(x^2-5x))^2=6^

$\Rightarrow {x}^{2} - 5 x = 36$

"rearrange into "color(blue)"standard form";ax^2+bx+c=0

$\text{subtract 36 from both sides}$

$\Rightarrow {x}^{2} - 5 x - 36 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{the factors of - 36 which sum to - 5 are - 9 and + 4}$

$\Rightarrow \left(x - 9\right) \left(x + 4\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x + 4 = 0 \Rightarrow x = - 4$

$x - 9 = 0 \Rightarrow x = 9$

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

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

$x = - 4 \to \sqrt{16 + 20} - 6 = \sqrt{36} - 6 = 6 - 6 = 0$

$x = 9 \to \sqrt{81 - 45} - 6 = \sqrt{36} - 6 = 6 - 6 = 0$

$\Rightarrow x = - 4 \text{ or "x=9" are the solutions}$