How do you find the roots, real and imaginary, of $y=(2x-8)(x-2)-x^2+4x$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y=(2x-8)(x-2)-x^2+4x$ using the quadratic formula?

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Answer 1 · 2017-05-07T08:20:02

The roots are $x=4" "or" "x=5/2$

Explanation:

$y=(2x-8)(x-2)-x^2+4x$
$" "$
$y=(2x-8)(x-2)-x(x-4)$
$" "$
$y=(2xxx-2xx4)(x-2)-(x-4)$
$" "$
$y=(2(x-4))(x-2)-(x-4)$
$" "$
$y=2(x-4)(x-2)-(x-4)$
$" "$
$y=(x-4)[2(x-2)-1]$
$" "$
$y=(x-4)[2x-4-1]$
$" "$
$y=(x-4)(2x-5)$
$" "$
We will find the roots when $y=0$
$" "$
$y=0$
$" "$
$rArr(x-4)(2x-5)=0$
$" "$
$x-4=0" "rArrx=4$
$" "$
$" "$ OR
$2x-5=0rArr2x=5rArrx=5/2$
$" "$
The roots are $x=4" "or" "x=5/2$

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How do you find the roots, real and imaginary, of $y=(2x-8)(x-2)-x^2+4x$ using the quadratic formula?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y=(2x-8)(x-2)-x^2+4x y=(2x-8)(x-2)-x^2+4x 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 $y=(2x-8)(x-2)-x^2+4x$ using the quadratic formula?