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

Mar 7, 2016

Equation has complex conjugate roots $1 + \frac{\sqrt{10}}{2} i$
and $1 - \frac{\sqrt{10}}{2} i$

#### Explanation:

To find the roots of −x^2−3−(x−2)^2=0, let us first simplify it.

−x^2−3−(x−2)^2=0 gives us

−x^2−3−(x^2−4x+4)=0 or

−x^2−3−x^2+4x-4=0 or

−2x^2+4x−7=0

Quadratic formula which gives solution of $a {x}^{2} + b x + c = 0$
is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

In the equation -2x^2+4x–7=0, $a = - 2$, $b = 4$ and $c = - 7$ and hence solution is

$x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \times \left(- 2\right) \times \left(- 7\right)}}{2 \times \left(- 2\right)}$

or $x = \frac{- 4 \pm \sqrt{16 - 56}}{- 4} = \frac{4}{4} \pm \frac{\sqrt{- 40}}{4}$

i.e. $x = 4 \pm 2 i \frac{\sqrt{10}}{4}$ or $x = 1 \pm \frac{\sqrt{10}}{2} i$

Hence equation has complex conjugate roots $1 + \frac{\sqrt{10}}{2} i$
and $1 - \frac{\sqrt{10}}{2} i$