What are the roots of this problem?

Question

i need more help!?! lol. as u have probably guessed, i have some really hard math hw. once again, its multiple choice.

y=(x^2 / 6) + x + 1/3

answers???

i need more help!?! lol. as u have probably guessed, i have some really hard math hw. once again, its multiple choice.

y=(x^2 / 6) + x + 1/3

i have to find the roots of the problem, but i ended up getting a really wrong answer.
the choices are

x= (-3 +/- sqrt of 11) / 2

x= -3 +/- sqrt of 11

x= (-3 +/- sqrt of 7) / 2

x= -3 +/- sqrt of 7

i really dont get how to solve it. answers??? and pleez explain.

Answer 1

finding roots of an equation means finding the solution of the equation when y = 0
x^2/6 + x + 1/3 = 0

multiply everything by 6 to get rid of denominator:
x^2 + 6x + 2 = 0

you can complete the square of use quadratic formula:

x^2 + 6x = -2
x^2 + 6x + (6/2)^2 = -2 + (6/2)^2
x^2 + 6x + 9 = -2 + 9
x^2 + 6x + 9 = 7

factor left side:
(x + 3)^2 = 7

take sq. rt:
x + 3 = +/- sq rt(7)
x = -3 +/- sq rt(7)

Answer 2

What's the question? Also, no choices are given!

Answer 3

What choices do you have ?