How do you solve #sqrt(x-3) = x - 5# and find any extraneous solutions?

Question

9539 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-29T08:28:18

#x = 7 or x = 4#

Both solutions are valid.

The value under the root may not be negative. #x>= 3#

Square both sides of the equation.
#sqrt(x-3)^2 = (x - 5)^2#

A common error is to square each of the terms on the right side separately, instead of as a square of a binomial.

#x-3 = x^2 -10x+25#

#x^2 -10x+25-x+3 = 0#

#x^2 -11x +28 = 0#

Find factors of 28 which add to give 11. (#7xx4# works)
The signs will be the same, both negative.

#(x-7)(x-4) = 0#

Putting each factor equal to 0 gives:

#x = 7 or x = 4#

Both solutions are valid.