# How do you solve sqrt( x -10) = 5x?

May 31, 2016

There are no Real solutions,
if Complex values are allowed $x = \frac{1 \pm i \sqrt{999}}{50}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \sqrt{x - 10} = 5 x$

Working in the domain of real numbers we must ensure that

$x \ge 10$ and $x \ge 0$ hence finally $x \ge 10$.

Squaring both sides:
$\textcolor{w h i t e}{\text{XXX}} x - 10 = 25 {x}^{2}$

$\textcolor{w h i t e}{\text{XXX}} 25 {x}^{2} - x + 10 = 0$

$\textcolor{w h i t e}{\text{XXX}} x = \frac{1 \pm \sqrt{- 999}}{50}$

Because Real numbers do not permit square roots of negative values there are no Real solutions.

May 31, 2016

We have complex roots only which are $x = \frac{1}{50} \pm i \frac{\sqrt{999}}{50}$

#### Explanation:

Squaring the two sides in $\sqrt{x - 10} = 5 x$, we get

$x - 10 = 25 {x}^{2}$

or $25 {x}^{2} - x + 10 = 0$

according to quadratic formula, the solution of the

equation $a {x}^{2} + b x + c = 0$ is given by

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

Hence, solution is $x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \cdot 25 \cdot 10}}{2 \cdot 25}$

or $x = \frac{1 \pm \sqrt{1 - 1000}}{50} = \frac{1 \pm \sqrt{- 999}}{50}$

As discriminant is negative, we only have complex roots,

which are $x = \frac{1 \pm i \sqrt{999}}{50}$