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\[\sqrt{\sqrt{x-2}+x}=2\]

Square both sides

root(x-2) + x =4
root(x-2) = 4-x
square again

x-2 = 16 - 8x + x^2
9x = 18 + x^2
x^2 - 9x + 18 = 0

(x-6)(x-3)=0
x=6 or x=3

@niravshah08 how on earth did you get the first line x-2 = 16 - 8x + x^2 ???

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okay i understand.... but is it wrong if i do this.....

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and if it is wrong please explain why

but there cant be two different ways of doing it... i think yours is right... hmm

Michelle, your method is wrong, because:
\[(a + b)^2 \neq a^2 + b^2\]

so what your saying is that i can do this:

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haha yeah, it gets a bit messy.. it's a pleasure :)