3 √x = 6

√x = 2 {divided each side by 3}

x = 4 {squared each side}

Check back in:

3 √4 = 6 {substituted 4 back in for x}

3(2) = 6 {evaluated the square root of 4}

6 = 6 {multiplied}

It works.

There are no extraneous solutions here. An extraneous solution is a solution that you get by solving the equation, but when you check it back into the equation, it does not satisfy the equation (make a true statement).

*- Algebra House*
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