Question

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

Answer 1

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}`