# POLQ 2 | Lesson 3 | Try This! (Solving Radical Equations) Solutions

Solutions:

1. Five is  a solution to the equation.
\begin{align}\sqrt{2x-1}-5 &= -2\\\\ \sqrt{2x-1} &= 5-2\\\\\sqrt{2x-1} &= 3\\\\ (\sqrt{2x-1})^2 &= 3^2\\\\2x-1 &=9\\\\2x &=10\\\\x &= 5\end{align}
2. Five is not a solution to this equation because $$8\neq2$$.  So there is not solution to this problem.
\begin{align}\sqrt{2x-1}+5 &= 2\\\\\sqrt {2x-1} &= -5+2\\\\\sqrt{2x-1} &= -3\\\\sqrt {2x-1})^2 &= (-3)^2\\\\2x-1 &=9\\\\2x &=10\\\\x &=5\end{align} However, when plugging 5 back into the original equation: \begin{align} \sqrt{2(5)-1}+5 &=2 \\\\ \sqrt{9}+5 &=2 \\\\ 8&=2\end{align} Therefore, \(x=5 is not a solution because $$8\neq2$$

Always check to make sure the solution satisfies the equation.