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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)

User Balbino
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2 Answers

7 votes

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)
User Fred Novack
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7.1k points
1 vote

Answer:

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)

User Marek Klein
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