How do you find the roots, real and imaginary, of #y= 8x^2-15x-4-2(3x+3)^2 # using the quadratic formula?

1 Answer
Jan 12, 2016

Answer:

#x~~-4.83# or #x~~-0.27#

Explanation:

First multiply out the terms so that the expression can be rearranged into standard form.
#y = 8x^2 - 15x - 4 -2(9x^2+ 18x +9)#
#y = 8x^2 - 18x^2 -15x -36x - 4 - 9#
#y = -10x^2-51x -13#
For a quadratic expression #y=ax^2 + bx+c# the quadratic formula is #x=(-b+-sqrt(b^2-4ac))/(2a#
#x = (-(-51)+-sqrt((-51)^2 -4(-10)(-13)))/(2(-10)#
#x=(51+-sqrt(2601-520))/-20#
#x = -(51+-sqrt(2081))/20#
Because the number under the square root sign is positive, there are no imaginary roots, only real ones.
#x~~-(51+-45.62)/20#
#x~~-4.83# or #x~~-0.27#