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March 20th, 2016, 01:59 PM   #1
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quadratic equation stumped

Hey guys
I am stumped on the penultimate and final line of the attached. Any help would be greatly appreciated in trying to figure out how they got to this.
Thanks for the help
J
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March 20th, 2016, 02:37 PM   #2
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$15y^2-32y+7 = 0$

$a = 15$, $b=-32$, $c=7$

$y = \dfrac{32 \pm \sqrt{(-32)^2 - 4(15)(7)}}{30}$

$y = \dfrac{32 \pm \sqrt{1024 - 420}}{30}$

$y = \dfrac{32 \pm \sqrt{604}}{30}$

note ... $604 = 4 \cdot 151$ ...

$y = \dfrac{32 \pm \sqrt{4}\cdot\sqrt{151}}{30}$

$y = \dfrac{32 \pm 2\cdot\sqrt{151}}{30}$

$y = \dfrac{\cancel{2}(16 \pm \sqrt{151})}{\cancel{2} \cdot 15}$

$y = \dfrac{16 \pm \sqrt{151}}{15}$
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March 20th, 2016, 02:38 PM   #3
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In the second-last line, they substituted $y = \dfrac{32 \pm \sqrt{604}}{30}$ into the equation $x = 4y - 4$.

The last line is just simplification.
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March 20th, 2016, 02:57 PM   #4
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Hey Azzajazz
Can you walk me through the simplification steps?
Thanks
J
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March 20th, 2016, 03:38 PM   #5
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$y = \dfrac{32 \pm \sqrt{604}}{30}$

so $x = 4y - 4 = \dfrac{4\times32 \pm 4\sqrt{604}}{30} - \dfrac{4\times30}{30}$.

Now use 128 - 120 = 8.
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