Question

How do you find the zeros, real and imaginary, of `y= -9x^2-2x-3` using the quadratic formula?

Answer 1

Zeros are `1/9-isqrt26/9` and `1/9+isqrt26/9`, two complex conjugate numbers

Explanation:

Zeros of a quadratic function `y=f(x)=-9x^2-2x-3` are those values of `x` for which `y=0`.

Using quadratic formula for a function `f(x)=ax^2+bx+c`,

zeros are `(-b+-sqrt(b^2-4ac))/(2a)`

and hence for `y=f(x)=-9x^2-2x-3`,

zeros are `(-(-2)+-sqrt((-2)^2-4xx(-9)xx(-3)))/(2xx(-9))`

i.e. `(2+-sqrt(4-108))/(-18)=(2+-sqrt(-104))/18`

= `(2+-sqrt(2^2xx26xxi^2))/18` - as `i^2=-1`

= `(2+-i2sqrt26)/18=1/9+-isqrt26/9`

Hence zeros are `1/9-isqrt26/9` and `1/9+isqrt26/9`, two complex conjugate numbers