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Real Roots

f(x) = 4x2 + 12x + 9.
a) Determine the number of real roots that exist for the equation f(x) = 0.

as b2 = 4ac

144 = 9 * 4 * 4
I said two real roots

Why is there only one
Original post by jsmith6131
f(x) = 4x2 + 12x + 9.
a) Determine the number of real roots that exist for the equation f(x) = 0.

as b2 = 4ac

144 = 9 * 4 * 4
I said two real roots

Why is there only one

There is a single repeated root in the case b2−4ac=0b^2-4ac=0 (which is the case you've used). There would be two distinct real roots if b2−4ac>0b^2-4ac>0.
ok thanks
Reply 3
Avatar for nuodai
nuodai
17
If you ever get mixed up with the discriminant thing, look at the quadratic formula:

x=−b±b2−4ac2x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2}

If b2−4ac>0b^2-4ac > 0 then the square root after the ±\pm sign is defined (and greater than zero), so the two roots corresponding to the + and the - are distinct.

If b2−4ac=0b^2-4ac = 0 then the formula just becomes x=−b2x = \dfrac{-b}{2} -- i.e. there is only one solution.

If b2−4ac<0b^2-4ac < 0 then the square root isn't defined (for real numbers), so there are no real roots.

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