How do you find the zeros, real and imaginary, of y= -x^2-5x-14 using the quadratic formula?

1 Answer
Alan P.
Nov 22, 2015

Two imaginary roots (-(5+-sqrt(31)i)/2)

Explanation:

For a quadratic equation in the general form
color(white)("XXX")y=ax^2+bx+c
the quadratic formula
color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)
gives us the zeros (or "roots") of the equation.

For the specific equation
color(white)("XXX")y=-x^2-5x-14
color(white)("XXXXXXXXXXXX")a=(-1)
color(white)("XXXXXXXXXXXX")b=(-5)
color(white)("XXXXXXXXXXXX")c=(-14)
so the quadratic formula becomes
color(white)("XXX")x=(-(-5)+-sqrt((-5)^2-4(-1)(-14)))/(2(-1))

color(white)("XXX")=(5+-sqrt(25-56))/(-2)

color(white)("XXX")=-(5+-sqrt(31)i)/2

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