How do you solve #sqrt(x+5)=1-x #?

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Answer 1 · 2016-05-14T00:25:31

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

#(sqrt(x + 5))^2 = (1 - x)^2#

#x + 5 = 1 - 2x + x^2#

#0 = x^2 - 3x - 4#

#0 = (x - 4)(x + 1)#

#x = 4 and -1#

However, after checking in the original equation we find that only #x= -1# works.

The solution set is #{-1}#.

Hopefully this helps!