How do you find the roots, real and imaginary, of $y=-(x+5)^2+2x^2 + 5x - 12$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y=-(x+5)^2+2x^2 + 5x - 12$ using the quadratic formula?

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Answer 1 · 2017-07-11T15:21:23

$x_1~~6.437+0i$
$x_2~~-1.437+0i$

Explanation:

Quadratic formula, given by:
$(-b+-sqrt(b^2-4ac))/(2a)$

Before we proceed, let's us expand the equation of $y$
$y=-(x+5)^2+2x^2+5x-12$
$y=-(x^2+10x+25)+2x^2+5x-12$
$y=-x^2-10x-25+2x^2+5x-12$
$y=(2-1)x^2+(5-10)x+(-37)$
$y=(1)x^2+(-5)x+(-37)$

Determine the $a$ , $b$ , and $c$
$a=1$
$b=-5$
$c=-37$

Use the formula:
$x=(-b+-sqrt(b^2-4ac))/(2a)$
$x=(-(-5)+-sqrt((-5)^2-4(1)(-37)))/(2(1))$
$x=(5+-sqrt(25+37))/2$
$x=(5+-sqrt(62))/2$

$x_1=(5+sqrt(62))/2~~6.437$
$x_2=(5-sqrt(62))/2~~-1.437$

In this case, the roots are real, their imaginary parts are 0
$x_1~~6.437+0i$
$x_2~~-1.437+0i$

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How do you find the roots, real and imaginary, of $y=-(x+5)^2+2x^2 + 5x - 12$ using the quadratic formula?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y=-(x+5)^2+2x^2 + 5x - 12 y=-(x+5)^2+2x^2 + 5x - 12 using the quadratic formula?

1 Answer

Explanation:

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Impact of this question

How do you find the roots, real and imaginary, of $y=-(x+5)^2+2x^2 + 5x - 12$ using the quadratic formula?