Question

How do you find the zeros, real and imaginary, of `y=- 7x^2-15x -35` using the quadratic formula?

Answer 1

`x=(15+isqrt(755))/(-14),` `(15-isqrt(755))/(-14)`

Explanation:

The zeros of a quadratic equation are the values for `x`, and are where the parabola crosses the x-axis.

`y=-7x^2-15x-35`

Substitute a zero for the `y`.

`0=-7x^2-15x-35` is a quadratic equation in the form: `ax^2+bx+c=0`, where `a=-7`, `b=-15`, `c=-35`.

Use the quadratic formula to solve for `x` (the zeros).

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

Substitute the given values for `a, b, and c` into the formula.

`x=(-(-15)+-sqrt((-15)^2-4*-7*-35))/(2*-7)`

Simplify.

`x=(15+-sqrt(-755))/(-14)`

`x=(15+isqrt(755))/(-14),` `(15-isqrt(755))/(-14)`

Answer 2

2 imaginary roots:
`x = (-15 +- sqrt755)/14`

Explanation:

`y = - (7x^2 + 15x + 35) = 0`
Use new improved quadratic formula (Socratic Search)
`D = d^2 = b^2 - 4ac = 225 - 980 = - 755 = 755i^2` -->
`d = +- isqrt755`
Because D < 0, there are 2 imaginary roots.
`x = - b/(2a) +- d/(2a) = - 15/14 +- isqrt755/14 = (-15 +- isqrt755)/14`

Note. Using the improved quadratic formula, the proceeding is much simpler, and the numeric computation is much easier.