How do you find the roots, real and imaginary, of $y= -3x^2 -2x +(2x+1)^2$ using the quadratic formula?

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Answer 1 · 2018-07-24T04:01:57

$y=-3x^2-2x+(2x+1)^2$ has a double root at $x=-1$

$y=-3x^2-2x+(2x+1)^2$
$y=-3x^2-2x+(4x^2+4x+1)$
$y=-3x^2-2x+4x^2+4x+1$
$y=x^2+2x+1$

Using the quadratic formula
$x=(-2+-sqrt(4-4(1)(1)))/(2)$
$x=(-2+-sqrt0)/2$
$x=(-2)/2$
$x=-1$

Therefore, $y=-3x^2-2x+(2x+1)^2$ has a double root at $x=-1$

How do you find the roots, real and imaginary, of y= -3x^2 -2x +(2x+1)^2 y= -3x^2 -2x +(2x+1)^2 using the quadratic formula?