how do sove for the solutions for ; the square root of x+ the square root of x-36 which then =0?

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how do sove for the solutions for ; the square root of x+ the square root of x-36 which then =0?

Mathematics
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Subtract the square root of x-36 from both sides! now: the square root of x = -the square root of x-36
\[\sqrt{x} = -\sqrt{x-36}\]
what about the 2?

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Other answers:

Square both of them\[(\sqrt{x})^{2} = (-\sqrt{x-36})^{2}\]
what 2
?
im so sry the whole thing is equal to 2 not 0 im so sry
ok...then it is: \[(\sqrt{x})^{2} = (2-\sqrt{x-36})^{2}\]
\[x = 4 - 2\sqrt{x-36} - 2\sqrt{x-36} + x - 36\]
\[x = -32 + x + -4\sqrt{x-36}\]
\[x - x + 32 = -4\sqrt{x-36}\]
\[32 = -4\sqrt{x-36}\]
\[(32)^{2} = (-4\sqrt{x-36})^{2}\]
no no!!
wait a sec!
Instead of square rooting 32 and the other, first divide -4 from both sides!
\[(-8)^{2} = (\sqrt{x-36})^{2}\]
you following?
64 = X-36 64 + 36 = X - 36 + 36 100 = X
thank you! :)

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