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

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Answer 1 · 2018-02-19T21:47:04

The roots are:
$\frac{5 + \sqrt{17}}{- 4}, and \frac{5 - \sqrt{17}}{- 4}$ $(5+sqrt(17))/(-4), and(5-sqrt(17))/(-4)$

First, expand the equation:

$- 2 (x^{2} + 2 x + 1) - x + 1$ $-2(x^2+2x+1)-x+1$

$- 2 x^{2} - 4 x - 2 - x + 1$ $-2x^2-4x-2-x+1$

$- 2 x^{2} - 5 x - 1$ $-2x^2-5x-1$

Now that it is in $a x^{2} + b x + c$ $ax^2+bx+c$ standard form, with

$a = - 2, b = - 5, c = - 1$ $a=-2,b=-5,c=-1$ ,

you can plug into the quadratic formula:

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

$\frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 \cdot - 2 \cdot - 1}}{2 \cdot - 2}$ $(-(-5)+-sqrt((-5)^2-4*-2*-1))/(2*-2)$

$\frac{5 \pm \sqrt{17}}{- 4}$ $(5+-sqrt(17))/(-4)$

The answer is

$\frac{5 + \sqrt{17}}{- 4}, and \frac{5 - \sqrt{17}}{- 4}$ $(5+sqrt(17))/(-4), and(5-sqrt(17))/(-4)$

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