How do you solve $sqrt(x+4) = sqrt(3x)$ and find any extraneous solutions?

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Answer 1 · 2016-06-12T22:09:08

$sqrt(x + 4) = sqrt(3x)$

$(sqrt(x + 4))^2 = (sqrt(3x))^2$

$x + 4 = 3x$

$4 = 3x - x$

$4 = 2x$

$2 = x$

Checking back in the original equation, we have that $sqrt(6) = sqrt(6)$ , so the solution works.

Our solution set is ${x = 2}$ .

Hopefully this helps!

How do you solve sqrt(x+4) = sqrt(3x)sqrt(x+4) = sqrt(3x) and find any extraneous solutions?