# How do you solve 45-sqrt(10-2x^2)=25?

Jun 29, 2017

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{45}$ from each side of the equation to isolate the radical term while keeping the equation balanced:

$- \textcolor{red}{45} + 45 - \sqrt{10 - 2 {x}^{2}} = - \textcolor{red}{45} + 25$

$0 - \sqrt{10 - 2 {x}^{2}} = - 20$

$- \sqrt{10 - 2 {x}^{2}} = - 20$

Next, square both sides of the equation to eliminate the radical while keeping the equation balanced:

${\left(- \sqrt{10 - 2 {x}^{2}}\right)}^{2} = {\left(- 20\right)}^{2}$

$10 - 2 {x}^{2} = 400$

Then, subtract $\textcolor{red}{10}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:

$- \textcolor{red}{10} + 10 - 2 {x}^{2} = - \textcolor{red}{10} + 400$

$0 - 2 {x}^{2} = 390$

$- 2 {x}^{2} = 390$

Then, divide each side of the equation by $\textcolor{red}{- 2}$ to isolate ${x}^{2}$ while keeping the equation balanced:

$\left(- 2 {x}^{2}\right) \textcolor{red}{- 2} = \frac{390}{\textcolor{red}{- 2}}$

$\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} {x}^{2}\right) \cancel{\textcolor{red}{- 2}} = - 195$

${x}^{2} = - 195$

Because any number squared always produces a positive result, there is no solution for $x$ which will result in a negative 195.

Or, the solution is the null or empty set: $\left\{\emptyset\right\}$