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

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Answer 1 · 2016-04-22T16:00:34

Square both sides, solve the resulting quadratic, then check the solutions to find valid solution:

#x = 8#

First square both sides, noting that this may introduce spurious solutions, to get:

#x+1 = x^2-10x+25#

Subtract #x+1# from both sides to get:

#0 = x^2-11x+24 = (x-3)(x-8)#

So #x = 3# or #x = 8#

The solution #x=3# of this quadratic is spurious since #x-5 = -2 < 0#, so does not match the positive square root in the original equation.

The solution #x=8# is a valid solution of the original equation:

#sqrt(8+1) = sqrt(9) = 3 = 8 - 5#