Question

How do you find the zeros of `y = -2x^2 - 11x + 17` using the quadratic formula?

Answer 1

`x=-11/4+sqrt(257)/4`
`x=-11/4-sqrt(257)/4`

`xapprox1.258`
`xapprox-6.758`

Explanation:

Standard form of quadratic equation:
`ax^2+bx+c`

Therefore in this equation:
`a=-2`
`b=-11`
`c=17`

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

Plug in givens and solve:
`x=(11+-sqrt((-11)^2-4(-2)(17)))/(2*-2)`

`x=(11+-sqrt(121+136))/(-4)`

`x=(11+-sqrt(257))/(-4)`

`x=-11/4+sqrt(257)/4`
`x=-11/4-sqrt(257)/4`

`xapprox1.258`
`xapprox-6.758`

Graph to confirm:
graph{-2x^2-11x+17 [-10, 10, -5, 5]}

Answer 2

`x=-(11+-sqrt257)/4` or `x=-(11+sqrt257)/4, -(11-sqrt257)/4`

Explanation:

The quadratic formula is `(-b+-sqrt(b^2-4ac))/(2a)`

This equation is written in standard form: `y=ax^2+bx+c`

To find the zeros using the quadratic formula, let's first establish what a, b, and c are.

`y=-2x^2-11x+17 rarr` -2 is a, -11 is b, 17 is c

Plug these numbers into the formula:

`(-(-11)+-sqrt((-11)^2-4*-2*17))/(2*-2)`

`(11+-sqrt(121-4*-2*17))/(2*-2)`

`(11+-sqrt(121-(-136)))/-4`

`(11+-sqrt257)/-4`

`-(11+-sqrt257)/4 rarr` This is your final answer because the square root of 257 cannot be reduced

The answer to the equation is `x=-(11+-sqrt257)/4` or `x=-(11+sqrt257)/4, -(11-sqrt257)/4`. These are the zeros (when plugged in for x, they make the function zero).