How do you find the roots, real and imaginary, of $y=2x^2 + 13x + 6+4(x -1)^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y=2x^2 + 13x + 6+4(x -1)^2$ using the quadratic formula?

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Answer 1 · 2015-12-29T11:30:00

$x = (-5 + sqrt(-215))/12$ or $x = (-5 - sqrt(-215))/12$

Explanation:

First expand and simplify to ensure just one term for each power of x.
$y = 2x^2 +13x +6 +4(x^2 -2x + 1)$
$y = 2x^2 +13x + 6 + 4x^2 -8x +4$
$y=6x^2 +5x +10$
Then use $x = (-b +- sqrt(b^2 - 4ac))/(2a)$
$x = (-5 +- sqrt( 5^2 - 4*6*10))/(2*6)$
$x = (-5 +-sqrt(25 - 240))/12$
$x = (-5 +- sqrt(-215))/12$
There are only imaginary roots to this equation, meaning that in graphical terms the graph never intersects the x axis.

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How do you find the roots, real and imaginary, of $y=2x^2 + 13x + 6+4(x -1)^2$ using the quadratic formula?

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Science

Math

Humanities

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How do you find the roots, real and imaginary, of y=2x^2 + 13x + 6+4(x -1)^2 y=2x^2 + 13x + 6+4(x -1)^2 using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of $y=2x^2 + 13x + 6+4(x -1)^2$ using the quadratic formula?