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Solving Radical Equations

Solving Radical Equations 1.19 - 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!
2x13=0\sqrt{2x-1}-3=0
2x1=3\sqrt{2x-1}=3
2x1=92x-1=9
2x=102x=10
x=5x=5
Next, we will check for extraneous solutions. We do that by substituting 55 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.
2x13=0\sqrt{2x-1}-3=0
2(5)13=?0\sqrt{2({\color{#0000FF}{5}})-1}-3\stackrel{?}{=}0
Simplify left-hand side
1013=?0\sqrt{10-1}-3\stackrel{?}{=}0
93=?0\sqrt{9}-3\stackrel{?}{=}0
33=?03-3\stackrel{?}{=}0
0=0 0=0 \ {\color{#009600}{\huge{\checkmark}}}
Because our substitution produced a true statement, we know that our answer, x=5,x=5, is correct.