# How do you solve 5 sqrt (x-6) = x?

Jul 25, 2016

x = 10 , x = 15

#### Explanation:

Begin by dividing both sides of the equation by 5.

$\frac{{\cancel{5}}^{1} \sqrt{x - 6}}{\cancel{5}} ^ 1 = \frac{x}{5}$

$\Rightarrow \sqrt{x - 6} = \frac{x}{5}$

To obtain the value inside the square root, we $\textcolor{b l u e}{\text{square both sides}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sqrt{a} \times \sqrt{a} = {\left(\sqrt{a}\right)}^{2} = a} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

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

$\Rightarrow x - 6 = {x}^{2} / 25$

Now multiply both sides by 25 to eliminate the fraction.

$\Rightarrow 25 \left(x - 6\right) = {\cancel{25}}^{1} \times \frac{{x}^{2}}{\cancel{25}} ^ 1 \Rightarrow 25 \left(x - 6\right) = {x}^{2}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

The standard form of a quadratic equation is.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{a {x}^{2} + b x + c = 0} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Rearrange $25 \left(x - 6\right) = {x}^{2} \text{ into standard form}$

$\Rightarrow 25 x - 150 = {x}^{2} \Rightarrow {x}^{2} - 25 x + 150 = 0$

To factorise we consider the product of the factors of 150 which also sum to -25 . These are -10 and -15

$\Rightarrow \left(x - 10\right) \left(x - 15\right) = 0 \text{ and solving gives}$

$x = 10 , x = 15$