How do you solve #sqrt(x-3) = x - 5# and find any extraneous solutions?
1 Answer
Aug 29, 2016
Both solutions are valid.
Explanation:
The value under the root may not be negative.
Square both sides of the equation.
A common error is to square each of the terms on the right side separately, instead of as a square of a binomial.
Find factors of 28 which add to give 11. (
The signs will be the same, both negative.
Putting each factor equal to 0 gives:
Both solutions are valid.