How do you solve #sqrt(x + 7) = x - 5#?

1 Answer
Sep 7, 2016

Answer:

#x=9#

Explanation:

Given:

#sqrt(x+7) = x-5#

Square both sides of the equation to get:

#x+7 = (x-5)^2 = x^2-10x+25#

Subtracting #x+7# from both ends, we get:

#0 = x^2-11x-18 = (x-2)(x-9)#

So the solutions of this derived quadratic equation are #x=2# and #x=9#.

Any solutions of the original equation must be solutions of the derived quadratic equation, so #x=2# and #x=9# are the only possible solutions of the original equation.

However, note that squaring is not a one to one function, so solutions of the derived equation may not be solutions of the original one.

In fact, we find:

#sqrt(color(red)(2)+7) = sqrt(9) = 3 != -3 = color(red)(2)-5#

So #x=2# is not a solution of the original equation.

On the other hand, we find:

#sqrt(color(red)(9)+7) = sqrt(16) = 4 = color(red)(9)-5#

So #x=9# is a solution of the original equation.