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 Algebra Pre-Algebra and Basic Algebra Math Forum

March 20th, 2016, 01:59 PM   #1
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Joined: Mar 2016
From: London

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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
Attached Images algebra query.jpg (9.3 KB, 23 views) 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. Tags equation, quadratic, stumped Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post HelpMeNow Algebra 7 November 3rd, 2014 05:00 AM jiasyuen Elementary Math 1 January 19th, 2014 01:03 PM mich89 Algebra 3 January 9th, 2013 01:22 PM mmcdonald Calculus 3 March 1st, 2012 05:28 AM STARGIRL Algebra 1 August 24th, 2008 01:04 PM

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