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

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

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Answer 1 · 2018-01-26T15:55:05

Imaginary Roots , $\frac{- 1 \pm \sqrt{23} i}{- 6}$ $(-1+-sqrt(23)i)/(-6)$

Explanation:

Simply this equation to get : $y = a x^{2} + b x + c$ $y=ax^2+bx+c$

$y = - x^{2} - 3 x - 2 (x^{2} - 2 x + 1)$ $y=-x^2-3x-2(x^2-2x+1)$
$y = - x^{2} - 3 x - 2 x^{2} + 4 x - 2$ $y=-x^2-3x-2x^2+4x-2$

$y = - 3 x^{2} + x - 2$ $y=-3x^2+x-2$

comparing with: $y = a x^{2} + b x + c$ $y=ax^2+bx+c$

we get $a = - 3$ $a=-3$ , $b = 1$ $b=1$ , $c = - 2$ $c=-2$

using the quadratic formula:

$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$
$\frac{- 1 \pm \sqrt{1^{2} - 4 (- 3) (- 2)}}{2 (- 3)}$ $(-1+-sqrt(1^2-4(-3)(-2)))/(2(-3)$

Solve further and get
answer = $\frac{- 1 \pm \sqrt{23} i}{- 6}$ $(-1+-sqrt(23)i)/(-6)$

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

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Explanation:

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