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

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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.

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