# How do you solve sqrt(15-2x)=x?

Jul 23, 2015

You square both sides of the equation and solve the resulting quadratic equation.

#### Explanation:

The first thing you need to do in order to solve this equation is to get rid of the square root by squaring both sides of the equation.

This will get you

${\left(\sqrt{15 - 2 x}\right)}^{2} = {x}^{2}$

$15 - 2 x = {x}^{2}$

Next, move all the terms on one side of the equation

${x}^{2} + 2 x - 15 = 0$

You can easily solve this one by using the quadratic formula, which states that, for any quadratic equation that takes the form $a {x}^{2} + b x + c = 0$, the solutions for $x$ can be determined by

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In your case, $a = 1$, $b = 2$, and $c = - 15$.

This means that you get

${x}_{1 , 2} = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \cdot 1 \cdot \left(- 15\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{- 2 \pm \sqrt{64}}{2} = \frac{- 2 \pm 8}{2}$

The two solutions for $x$ will be

$\left\{\begin{matrix}{x}_{1} = \frac{- 2 + 8}{2} = 3 \\ {x}_{2} = \frac{- 2 - 8}{2} = - 5\end{matrix}\right.$

Check to see if the two solutions are valid

$\sqrt{15 - 2 \cdot 3} = 3 \iff \sqrt{9} = 3 \iff 3 = 3 \to \textcolor{g r e e n}{\text{valid}}$

and

$\sqrt{15 - 2 \cdot \left(- 5\right)} = - 5 \iff \sqrt{25} = - 5 \to \textcolor{red}{\text{not valid}}$

Therefore, the only solution to this equation is

$x = \textcolor{g r e e n}{3}$

Jul 23, 2015

Square both sides and then solve as a quadratic to obtain:
$\textcolor{w h i t e}{\text{XXXX}}$$x = 3$

#### Explanation:

$\sqrt{15 - 2 x} = x$

$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$15 - 2 x = {x}^{2}$

$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + 2 x - 15 = 0$

$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$\left(x + 5\right) \left(x - 3\right) = 0$

$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$x = - 5$$\textcolor{w h i t e}{\text{XXXX}}$or$\textcolor{w h i t e}{\text{XXXX}}$$x = 3$

Since $\sqrt{\text{anything}}$$\textcolor{w h i t e}{\text{XX}}$ is, by definition, positive
$\textcolor{w h i t e}{\text{XXXX}}$we can ignore $x = - 5$ as extraneous

Leaving $x = 3$ as the only valid solution.