How do you find the zeros, real and imaginary, of #y=-x^2-x-9# using the quadratic formula?

1 Answer
Mar 25, 2018

Answer:

#x=-1/2+- 1/2sqrt(35)color(white)(.)i#

Explanation:

You have two ways of tackling this. You may complete the square or use 'the formula'

I choose to use the formula. It really is worth trying to commit this one to memory. It crops up quite often.

Standard form #y=ax^2+bx+c color(white)("dd") -> color(white)("dd")x=(-b+-sqrt(b^2-4ac))/(2a)#

Given #y=-x^2-x-9#

where #a=-1;b=-1 and c=-9#

#x=(-(-1)+-sqrt((-1)^2-4(-1)(-9)))/(2(-1))#

#x=(+1+-sqrt(-35))/(-2)#

#x=-1/2+-(sqrt(35)xxsqrt(-1))/2#

#x=-1/2+- 1/2sqrt(35)color(white)(.)i#