How do you find the zeros, real and imaginary, of #y= -2x^2-x-19 # using the quadratic formula?

1 Answer
May 29, 2016

Answer:

Zeros of #f(x)# are #{-1/4+isqrt151/4,-1/4-isqrt151/4}#

Explanation:

To find zeros of #f(x)=-2x^2-x-19#, we need roots of the equation #-2x^2-x-19=0#, which can be obtained using quadratic formula.

As the roots of #ax^2+bx+c=0# are #x=(-b+-sqrt(b^2-4ac))/(2a)#

Hence, roots of zeros of #-2x^2-x-19=0# are

#x=(-(-1)+-sqrt((-1)^2-4(-2)(-19)))/(2(-2))#

= #(1+-sqrt(1-152))/(-4)=-1/4+-isqrt151/4#

Zeros of #f(x)# are #{-1/4+isqrt151/4,-1/4-isqrt151/4}#