Question

How do you find the zeros, real and imaginary, of `y=-4x^2-4x-15` using the quadratic formula?

Answer 1

Finding the values of `x` for `color(white)(xxx)y=-4x^2-4x-15 = 0`

`x!in RRcolor(white)(xx)` but `color(white)(xx) x inCC -> x=-1/2 +-isqrt(7/2)`

Explanation:

Using standard for of `y=ax^2+bx+c`

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

Let:
`a=-4`
`b=-4`
`c=-15`

So by substitution we have:

`x= (-(-4) +- sqrt((-4)^2 -(4)(-4)(-15)))/(2(-4)`

`x=(+4 +- sqrt(16-240))/(2(-4))`

`x=(4+-sqrt(-224))/(-8)`

partitioning 224 into prime numbers and squaring where able

`x= (4+- sqrt((-1) times 2^2 times 2^2 times 2 times 7))/(-8)`

`x= (4+-4sqrt(-14))/(-8)`

`x=(1+- i sqrt(14))/(-2)`
As we have `+-` in the numerator before the root having a negative denominator makes no tangible difference to that root. However, it does have an effect on the 1 preceding it. Consequently we have `-1/2`.

`x=-1/2 +- isqrt(14/4)`

`x=-1/2 +-isqrt(7/2)`