How do you solve $sqrt(x)= x-2$ ?

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Answer 1 · 2016-06-29T15:34:19

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

$x = x^2 - 4x + 4$

$0 = x^2 - 5x + 4$

$0 = (x - 4)(x - 1)$

$x = 4 and 1$

Checking in the original equation, we find that only $x = 4$ works.

The solution set is ${4}$

Hopefully this helps!

How do you solve sqrt(x)= x-2sqrt(x)= x-2?