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

1 Answer
Lil Del
Oct 17, 2016

The roots (both real) are #x = 3 + sqrt3# and #x = 3 - sqrt3#.

Explanation:

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

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

For the quadratic function #y = -x^2 - 6x - 6#, #a = -1#, #b = -6#, and #c = -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 +- sqrt(36 - 24))/-2#
#x = (6 +- sqrt(12))/-2#
#x = (-6 +- 2sqrt3)/-2#
#x = 3 +- sqrt3#

So, the roots of #y = -x ^2 -6x -6# are #x = 3 + sqrt3# and #x = 3 - sqrt3#.

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