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Solving Radical Equations

Solving Radical Equations 1.15 - Solution

arrow_back Return to Solving Radical Equations
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!
4x233=2\sqrt{4x-23}-3=2
4x23=5\sqrt{4x-23}=5
4x23=254x-23=25
4x=484x=48
x=12x=12
Next, we will check for extraneous solutions. We do that by substituting 1212 for xx 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.
4x233=2\sqrt{4x-23}-3=2
4(12)233=?2\sqrt{4({\color{#0000FF}{12}})-23}-3\stackrel{?}{=}2
Simplify left-hand side
48233=?2\sqrt{48-23}-3\stackrel{?}{=}2
253=?2\sqrt{25}-3\stackrel{?}{=}2
53=?25-3\stackrel{?}{=}2
2=2 2=2 \ {\color{#009600}{\huge{\checkmark}}}
Because our substitution produced a true statement, we know that our answer, x=12,x=12, is correct.