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!