How do you solve #sqrt(2x – 1) = x – 2#?

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Answer 1 · 2016-06-15T03:40:05

Start by squaring both sides.

#(sqrt(2x - 1))^2 = (x - 2)^2#

#2x - 1 = x^2 - 4x+ 4#

#0 = x^2 - 6x + 5#

#0 = (x - 5)(x - 1)#

#x = 5 and 1#

Checking in the original equation, we find that only #x = 5# works. Our solution set is therefore #{x = 5}#.

Hopefully this helps!