How do you find the roots, real and imaginary, of #y=-(2x-1)^2 -2x^2 - 3x + 4 # using the quadratic formula?

1 Answer
Dec 18, 2015

Answer:

#x=(1+-sqrt73)/12#

Explanation:

First, get the equation into the standard form of a quadratic equation

#y=ax^2+bx+c#

which can be very easily applied to the quadratic equation

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

To get the equation into standard form, simplify by combining like terms.

First, square and expand the #2x-1# term.

#y=-(2x-1)(2x-1)-2x^2-3x+4#

#y=-(4x^2-4x+1)-2x^2-3x+4#

#y=-4x^2+4x-1-2x^2-3x+4#

#y=-6x^2+x+3#

Thus,

#a=-6#
#b=1#
#c=3#

so,

#x=(-1+-sqrt(1^2-4(-6)(3)))/(2(-6))#

#x=(-1+-sqrt(1+72))/(-12#

#x=(1+-sqrt73)/12#