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

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Answer 1 · 2017-11-05T03:31:55

$y$ has two complex roots: $x=1/4(-1+- sqrt31 i)$

$y=x^2-3x+(-x-2)^2$

The roots of $y$ occur where $y=0$

$:. x^2-3x+(-x-2)^2 =0$

Expanding:

$x^2-3x+x^2+4x+4=0$

Combining like terms:

$2x^2+x+4=0$

$x= (-1+-sqrt((1)^2 - 4xx2xx4))/(2xx2)$

$= (-1+-sqrt(1-32))/4$

$= (-1+-sqrt(-31))/4$

$= (-1+-sqrt(31xx(-1)))/4$

$=1/4(-1+- sqrt31 i)$

$y$ has two complex roots: $x=1/4(-1+- sqrt31 i)$

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