## anonymous 5 years ago is there a value for h that makes it possible for the equation √(x+h)+5=0 to have any real number solution? explain

1. anonymous

is x+h+5 all under the square root? or just x+h?

2. anonymous

just x+h

3. anonymous

ok.. so you want to find a value for h that makes the expression $\sqrt{x+h}+5=0 \implies \sqrt{x+h}=-5$ but the square root can't have a negative value. so, such h doesn't exist.

4. anonymous

wait, the sqroot of a number can have a negative value, the SQUARE of a number can't have a real value

5. anonymous

square both sides and you get x+h = 25

6. anonymous

that's not right mindedone, you're suggesting that h=25-x is a solution.. check by plugging back into the equation and see what you get.

7. anonymous

Keep in mind though that if you plug h=25-x into the original equation you get +-sqrt(25) = -5, and so the negative square root does satisfy the equation.

8. anonymous

you get sqrt(25) = -5

9. anonymous

sqrt(25) = +-5