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

1 Answer
Nov 22, 2015

Answer:

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#