How do you solve $sqrt(33+sqrtx)=6$ ?

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Answer 1 · 2016-11-26T22:21:49

$x = 9$

Square both sides:

$(sqrt(33 + sqrt(x)))^2 = 6^2$

$33 + sqrt(x) = 36$

$sqrt(x) = 3$

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

$x = 9$

Hopefully this helps!

How do you solve sqrt(33+sqrtx)=6sqrt(33+sqrtx)=6?