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 $a=-1, b=-9, c=8$

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

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

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

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

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