How do you solve #sqrt(4x+5)=x#?

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Answer 1 · 2016-08-02T03:34:00

#(sqrt(4x + 5))^2 = x^2#

#4x + 5 = x^2#

#0 = x^2 - 4x - 5#

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

#x = 5 and -1#

However, checking in the original equation, we find that only #x = 5# works. Hence, the solution set is #{5}#.

Hopefully this helps!