Question

How do you find the roots, real and imaginary, of `y= -x^2-6x -6` using the quadratic formula?

Answer 1

The roots (both real) are `x = 3 + sqrt3` and `x = 3 - sqrt3`.

Explanation:

A quadratic function in standard form (`y = ax^2 + bx + c`) can be solved using its related quadratic equation (`ax^2 + bx + c = 0`) and the quadratic formula. The quadratic formula is:

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

For the quadratic function `y = -x^2 - 6x - 6`, `a = -1`, `b = -6`, and `c = -6`. Substitute these values into the quadratic formula and simplify following the Order of Operations.

`x = (-(-6) +- sqrt((-6)^2 - 4(-1)(-6)))/((2)(-1))`
`x = (6 +- sqrt(36 - 24))/-2`
`x = (6 +- sqrt(12))/-2`
`x = (-6 +- 2sqrt3)/-2`
`x = 3 +- sqrt3`

So, the roots of `y = -x ^2 -6x -6` are `x = 3 + sqrt3` and `x = 3 - sqrt3`.