How do you find the roots, real and imaginary, of $y = - x^{2} - x - 6 - {(x - 3)}^{2}$ $y= -x^2 -x-6-(x- 3 )^2$ using the quadratic formula?

Question

1692 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-15T05:54:26

$x = \frac{- 5 + i \sqrt{95}}{- 4}$ $x=(-5+isqrt95)/-4$ and
$x = \frac{- 5 - i \sqrt{95}}{- 4}$ $x=(-5-isqrt95)/-4$

First, you want to fully expand the equation then combine all like terms.
$- x^{2} - x - 6 - {(x - 3)}^{2} = 0$ $-x^2-x-6-(x-3)^2=0$

$- x^{2} - x - 6 - x^{2} + 6 x - 9 = 0$ $-x^2-x-6-x^2+6x-9=0$

$- 2 x^{2} + 5 x - 15 = 0$ $-2x^2+5x-15=0$

Now the equation is simplified to be able to be used in the quadratic formula.

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

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$(- 2) x^{2} + (5) x + (- 15) = 0$ $(-2)x^2+(5)x+(-15)=0$

$x = \frac{- 5 \pm \sqrt{5^{2} - (4 \cdot - 2 \cdot - 15)}}{2 \cdot - 2}$ $x=(-5+-sqrt(5^2-(4*-2*-15)))/(2*-2)$

$x = \frac{- 5 \pm \sqrt{- 95}}{- 4}$ $x=(-5+-sqrt(-95))/-4$

$x = \frac{- 5 \pm i \sqrt{95}}{- 4}$ $x=(-5+-isqrt95)/-4$

How do you find the roots, real and imaginary, of y=−x2−x−6−(x−3)2y= -x^2 -x-6-(x- 3 )^2 using the quadratic formula?