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

May 7, 2017

The roots are $x = 4 \text{ "or" } x = \frac{5}{2}$

#### Explanation:

$y = \left(2 x - 8\right) \left(x - 2\right) - {x}^{2} + 4 x$
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$y = \left(2 x - 8\right) \left(x - 2\right) - x \left(x - 4\right)$
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$y = \left(2 \times x - 2 \times 4\right) \left(x - 2\right) - \left(x - 4\right)$
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$y = \left(2 \left(x - 4\right)\right) \left(x - 2\right) - \left(x - 4\right)$
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$y = 2 \left(x - 4\right) \left(x - 2\right) - \left(x - 4\right)$
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$y = \left(x - 4\right) \left[2 \left(x - 2\right) - 1\right]$
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$y = \left(x - 4\right) \left[2 x - 4 - 1\right]$
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$y = \left(x - 4\right) \left(2 x - 5\right)$
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We will find the roots when $y = 0$
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$y = 0$
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$\Rightarrow \left(x - 4\right) \left(2 x - 5\right) = 0$
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$x - 4 = 0 \text{ } \Rightarrow x = 4$
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$\text{ }$OR
$2 x - 5 = 0 \Rightarrow 2 x = 5 \Rightarrow x = \frac{5}{2}$
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The roots are $x = 4 \text{ "or" } x = \frac{5}{2}$