My Math Forum quadratic equation stumped

 Algebra Pre-Algebra and Basic Algebra Math Forum

March 20th, 2016, 01:59 PM   #1
Newbie

Joined: Mar 2016
From: London

Posts: 6
Thanks: 0

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 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 $15y^2-32y+7 = 0$ $a = 15$, $b=-32$, $c=7$ $y = \dfrac{32 \pm \sqrt{(-32)^2 - 4(15)(7)}}{30}$ $y = \dfrac{32 \pm \sqrt{1024 - 420}}{30}$ $y = \dfrac{32 \pm \sqrt{604}}{30}$ note ... $604 = 4 \cdot 151$ ... $y = \dfrac{32 \pm \sqrt{4}\cdot\sqrt{151}}{30}$ $y = \dfrac{32 \pm 2\cdot\sqrt{151}}{30}$ $y = \dfrac{\cancel{2}(16 \pm \sqrt{151})}{\cancel{2} \cdot 15}$ $y = \dfrac{16 \pm \sqrt{151}}{15}$
 March 20th, 2016, 02:38 PM #3 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 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.
 March 20th, 2016, 02:57 PM #4 Newbie   Joined: Mar 2016 From: London Posts: 6 Thanks: 0 Hey Azzajazz Can you walk me through the simplification steps? Thanks J
 March 20th, 2016, 03:38 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 $y = \dfrac{32 \pm \sqrt{604}}{30}$ 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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