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

1 Answer
Jan 23, 2016

Answer:

#x=(1+-sqrt37)/2#

Explanation:

The quadratic formula find the zeroes of a quadratic equation in the form #y=ax^2+bx+c# through

#x=(-b+-sqrt(b^2-4ac))/(2a)#

In the given quadratic equation, we know that #a=1,b=-1,c=-9#. Plugging these into the quadratic formula gives

#x=(-(-1)+-sqrt((-1)^2-(4xx-9xx1)))/(2xx1)#

#x=(1+-sqrt(1-(-36)))/2#

#x=(1+-sqrt37)/2#

Thus, the two times when the function equal #0# are at #x=(1+sqrt37)/2# and #x=(1-sqrt37)/2#.