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

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Answer 1 · 2016-04-23T22:51:46

#x=4#

We isolate the square root first so that we can simplify by squaring.

First subtract #1# from both sides to get:

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

Next square both sides (which may result in spurious solutions) to get:

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

Subtract #x+5# from both sides to get:

#0 = x^2-3x-4 = (x-4)(x+1)#

Hence #x = 4# or #x = -1#

The solution #x=-1# of this quadratic is not a solution of the original equation:

#sqrt(-1+5) = sqrt(4) = 2 != -2 = -1-1#

The other solution #x=4# is a solution of the original equation:

#sqrt(4+5) = sqrt(9) = 3 = 4 - 1#