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

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Answer 1 · 2016-03-07T16:20:08

Equation has complex conjugate roots $1+sqrt10/2i$
and $1-sqrt10/2i$

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 $ax^2+bx+c=0$
is $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

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

$x=(-4+-sqrt(4^2-4xx(-2)xx(-7)))/(2xx(-2))$

or $x=(-4+-sqrt(16-56))/(-4)=4/4+-sqrt(-40)/4$

i.e. $x=4+-2isqrt10/4$ or $x=1+-sqrt10/2i$

Hence equation has complex conjugate roots $1+sqrt10/2i$
and $1-sqrt10/2i$

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