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March 20th, 2016, 01:59 PM  #1 
Newbie Joined: Mar 2016 From: London Posts: 6 Thanks: 0  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 
March 20th, 2016, 02:37 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,885 Thanks: 1504  $15y^232y+7 = 0$ $a = 15$, $b=32$, $c=7$ $y = \dfrac{32 \pm \sqrt{(32)^2  4(15)(7)}}{30}$ $y = \dfrac{32 \pm \sqrt{1024  420}}{30}$ $y = \dfrac{32 \pm \sqrt{604}}{30}$ 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 secondlast 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: 20,472 Thanks: 2039 
$y = \dfrac{32 \pm \sqrt{604}}{30}$ 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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