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

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Answer 1 · 2017-04-18T14:22:11

# x = (9 + sqrt(113)) / (-2) ~=9.8 and x = (9 - sqrt(113)) / (-2) ~=-0.8#

Let's first simplify this so that we can have it in the form of #y=ax^2+bx+c#:

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

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

#y=-3x^2 -x+2x^2-8x+8#

#y=-x^2 -9x+8#

Now we're ready for the quadratic formula:

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

and we have #a=-1, b=-9, c=8#

# x = (9 \pm sqrt((-9)^2-4(-1)(8))) / (2(-1)) #

# x = (9 \pm sqrt(81+32)) / (-2) #

# x = (9 \pm sqrt(113)) / (-2) #

# x = (9 + sqrt(113)) / (-2) ~=9.8 and x = (9 - sqrt(113)) / (-2) ~=-0.8#