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Im having trouble finding the zeros of this polynomial:
-x^2+2x+4
I have tried factoring it and it dosent work i have tried the quadratic formula too but no luck. I am supposed to graph this but I need the zeros first. Anyone have any suggestions im really lost. Thanks for your help in advance.
PS. The first term is read as "negative x squared" im not sure how to do the symbols...
$\displaystyle -x^2+2x+4 = 0$Quote:
Originally Posted by nascar77
$\displaystyle a = -1$
$\displaystyle b = 2$
$\displaystyle c = 4$
$\displaystyle x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(4)}}{2(-1)}$
now ... what exactly is your problem in evaluating the above expression that uses the quadratic formula?
Because the quadratic formula produces zeros that have imaginary numbers. But im actually supposed to graph this on a coordinate plane. The original funcion was f(x)=-x^3+4x^2-8. From here i used synthetic division and found one zero to be two. The quadratic i posted was the quotient left from the division.
is that so?Quote:
Because the quadratic formula produces zeros that have imaginary numbers.
$\displaystyle x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(4)}}{2(-1)}$
$\displaystyle x = \frac{-2 \pm \sqrt{4 - (-16)}}{-2}$
$\displaystyle x = \frac{-2 \pm \sqrt{20}}{-2}$
$\displaystyle x = \frac{-2 \pm 2\sqrt{5}}{-2}$
$\displaystyle x = 1 \mp \sqrt{5}$
I'd say that x has two real zeros.
I feel stupid I must have gotten lost somewhere in the algebra. thanks so much for your help!
-Kimberly
