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

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

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Answer 1 · 2018-07-24T04:01:57

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

Explanation:

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

Using the quadratic formula
$x = \frac{- 2 \pm \sqrt{4 - 4 (1) (1)}}{2}$ $x=(-2+-sqrt(4-4(1)(1)))/(2)$
$x = \frac{- 2 \pm \sqrt{0}}{2}$ $x=(-2+-sqrt0)/2$
$x = \frac{- 2}{2}$ $x=(-2)/2$
$x = - 1$ $x=-1$

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

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

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of y=−3x2−2x+(2x+1)2y= -3x^2 -2x +(2x+1)^2 using the quadratic formula?

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