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

2 Answers
Mar 24, 2018

Answer:

#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]}

Mar 24, 2018

Answer:

#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).