Question

what are the x-intercepts of the function f(x) = –2x2 – 3x 20? (–4, 0) and [5/2,0] [-5/2,0] and (4, 0) (–5, 0) and (2, 0) (–2, 0) and (5, 0)

Answer 1

`x= \frac{3 \pm \sqrt{ (-3)^(2) -4 * -2 * 20} }{2 * -2} \\ =(3 \pm √( 9 + 160) )/(-4) \\ =(3 \pm √( 169) )/(-4) \\ = (3+13)/(-4) \ or \ (3-13)/(-4) \\ =-4 \ or \ 2.5`
x-intercepts is (-4, 0) and (5/2, 0)

Answer 2

the x-intercepts are

`(-4,0)` and
`(5/2,0)`

Explanation:

we know that

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`f(x)=-2x^(2) -3x+20`

Equate the quadratic equation to zero to find the x-intercepts

`-2x^(2) -3x+20=0`

so

`a=-2\\b=-3\\c=20`

substitute in the formula

`x=\frac{3(+/-)\sqrt{(-3)^(2)-4(-2)(20)}} {2(-2)}`

`x=\frac{3(+/-)√(9+160)} {-4}`

`x=\frac{3(+/-)13} {-4}`

`x=\frac{3+13} {-4}=-4`

`x=\frac{3-13} {-4}=5/2`

therefore

the x-intercepts are

`(-4,0)` and
`(5/2,0)`