How do you find the roots, real and imaginary, of y= -x^2-6x -6 y=x26x6 using the quadratic formula?

1 Answer
Oct 17, 2016

The roots (both real) are x = 3 + sqrt3x=3+3 and x = 3 - sqrt3x=33.

Explanation:

A quadratic function in standard form (y = ax^2 + bx + cy=ax2+bx+c) can be solved using its related quadratic equation (ax^2 + bx + c = 0ax2+bx+c=0) and the quadratic formula. The quadratic formula is:

x = (-b +-sqrt(b^2 - 4ac))/(2a)x=b±b24ac2a

For the quadratic function y = -x^2 - 6x - 6y=x26x6, a = -1a=1, b = -6b=6, and c = -6c=6. Substitute these values into the quadratic formula and simplify following the Order of Operations.

x = (-(-6) +- sqrt((-6)^2 - 4(-1)(-6)))/((2)(-1))x=(6)±(6)24(1)(6)(2)(1)
x = (6 +- sqrt(36 - 24))/-2x=6±36242
x = (6 +- sqrt(12))/-2x=6±122
x = (-6 +- 2sqrt3)/-2x=6±232
x = 3 +- sqrt3x=3±3

So, the roots of y = -x ^2 -6x -6y=x26x6 are x = 3 + sqrt3x=3+3 and x = 3 - sqrt3x=33.