Does sqrt(-3x - 2) = x have two extraneous roots?

Question

I have been assured that the roots x=-1 and x=-2 are both extraneous.. and I reason has been explained to me, but I can't wrap my brains around it. When I check the roots by substituting them back into the original equation
sqrt(1) = -1 and sqrt(4) = -2
they appear to be true statements. HELP!

so.......... for a radical to be a true function (one y-value for each x-value), you only consider the positive root?

When do you consider the negative root?

Answer 1

It's a technical issue. The math symbol √(-3x-2) means the positive root, even though there is a corresponding negative number whose square is the same thing. (This is so square root is a legitimate function with a single output, things would be messier if it sometimes gave two answers.)

Responding to your "additional details": Yes, that's exactly right. The way to designate the negative roots would be -√(-3x-2). Whenever you raise both sides of an equation to a power (like squaring when there's a square root), these extraneous roots can arise -- always plug back in to check, remember the convention that √ is always non-negative (could be 0).

Answer 2

Yes. The first responder's solution is correct, as far as it goes, but the square root designates (by convention) the positive square root, and both of the numeric solutions are negative. Hence, they aren't really solutions, and are called extraneous.

Answer 3

-3x - 2 = x^2
x^2 + 3x +2 = 0
(x + 2) (x +1) = 0

x = -2 or -1