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

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

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Answer 1 · 2015-12-29T02:13:31

$y=2(x+3)^2-(x-3)^2$
$implies y=2(x^2+6x+9)-(x^2-6x+9)$
$implies y=2x^2+12x+18-x^2+6x-9$
$implies y=x^2+18x+9$
For finding roots becomes #0 $.$ implies x^2+18x+9=0#

The quadratic formula is
If $ax^2+bx+c=0$ then $x=(-b+-sqrt(b^2-4ac))/(2a)$
Here $a=1$ $b=18$ and $c=9$

$implies x=(-18+-sqrt(18^2-4*1*9))/(2*1)$

$implies x=(-18+-sqrt(324-36))/(2)$

$implies x=(-18+-sqrt(288))/(2)$

$implies x=(-18+-12sqrt(2))/(2)$

$implies x=(2(-9+-6sqrt(2)))/(2)$

$implies x=(-9+-6sqrt(2))$

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

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