How do you find the zeroes for #y=9x^2+6x-1#?

1 Answer
Feb 13, 2017

Answer:

-0.805 or 0.138

Explanation:

Assuming #y=0# you can use the quadratic formula

#x=(-b±sqrt(b^2-4ac))/(2a)#

to find the roots of a quadratic.
A quadratic is defined as #ax^2+bx+c# so for your case you may plug in the respective values of #a#, #b#, and #c#:

#a=9#
#b=6#
#c=-1#

Then the quadratic formula would look like this:

#x=(-6±sqrt(6^2-4*9*-1))/(2*9)#

Due to the #±# sign we then have two variants for 2 answers (2 is also the degree of the polynomial which determines the amount of roots/answers of the polynomial)

#x=(-6+sqrt(6^2-4*9*-1))/(2*9)=0.138071187458#
#x=(-6-sqrt(6^2-4*9*-1))/(2*9)=-0.804737854124#

I rounded the answers to #x=-0.805, x=0.138#
To check your answer a graph can be utilized

graph{9x^2+6x-1 [-5.285, 4.58, -2.74, 2.193]}

:-)