How do you solve #sqrt(x + 7) = x - 5#?
#sqrt(x+7) = x-5#
Square both sides of the equation to get:
#x+7 = (x-5)^2 = x^2-10x+25#
#0 = x^2-11x-18 = (x-2)(x-9)#
So the solutions of this derived quadratic equation are
Any solutions of the original equation must be solutions of the derived quadratic equation, so
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#
On the other hand, we find:
#sqrt(color(red)(9)+7) = sqrt(16) = 4 = color(red)(9)-5#