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# Solving Radical Equations

## Solving Radical Equations 1.15 - Solution

To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!
$\sqrt{4x-23}-3=2$
$\sqrt{4x-23}=5$
$4x-23=25$
$4x=48$
$x=12$
Next, we will check for extraneous solutions. We do that by substituting $12$ for $x$ into the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution.
$\sqrt{4x-23}-3=2$
$\sqrt{4({\color{#0000FF}{12}})-23}-3\stackrel{?}{=}2$
Simplify left-hand side
$\sqrt{48-23}-3\stackrel{?}{=}2$
$\sqrt{25}-3\stackrel{?}{=}2$
$5-3\stackrel{?}{=}2$
$2=2 \ {\color{#009600}{\huge{\checkmark}}}$
Because our substitution produced a true statement, we know that our answer, $x=12,$ is correct.