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

2 Answers
Hriman
Mar 24, 2018

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]}

Serena D.
Mar 24, 2018

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).

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