Julie is solving the equation x^2 + 6x + 9 = 0 and notices that the discriminant b^2 - 4ac has a value of 0. This tells her that the equation has
A)
no real roots.
B)
one real root.
Eliminate
C)
two real roots.
D)
three real roots.

Question

Julie is solving the equation x^2 + 6x + 9 = 0 and notices that the discriminant b^2 - 4ac has a value of 0. This tells her that the equation has
A)
no real roots.	
B)
one real root.	
Eliminate
C)
two real roots.	
D)
three real roots.

accessdenied · Answer

If the discriminant is 0, look at the quadratic formula where the discriminant is 0
$$ x = \frac{-b \pm \sqrt{\color{goldenrod}{b^2 - 4ac}}}{2a} $$

accessdenied · Answer

If that part is 0, then the square root just cancels out, and whether you add or subtract 0, you get the same number.
So...
$$ x = -\! \frac{b}{2a} $$

anonymous · Answer

so it would be A right

anonymous · Answer

no real roots

accessdenied · Answer

Nope, x=-b/2a is a real value.  Just in general terms.
You have one real root, which has multiplicity 2.

accessdenied · Answer

You could also factor it to figure it out, its a perfect square trinomial that you can factor into (x+3)^2
 x=-3