Algebra Pre-Algebra and Basic Algebra Math Forum

 November 13th, 2018, 11:09 AM #1 Newbie   Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0 factorize fraction a^2-3a-5=0 then what is the result of: (a^4+25)/(a^2) answer is 19 but can not get there i tried rearranging the first equation and taking square of both sides but it didnt work November 13th, 2018, 11:51 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 You were on the right track, but did not go far enough. $a^2 - 3a - 5 = 0 \implies a^2 = 3a + 5 \implies$ $a^4 = 9a^2 + 30a + 25 = 9(3a + 5) + 30a + 25 = 57a + 70 \implies$ $a^4 + 25 = 57a + 95 = 19(3a + 5) \implies \dfrac{a^4 + 25}{a^2} = \dfrac{19(3a + 5)}{3a + 5} = 19.$ Thanks from topsquark November 13th, 2018, 01:37 PM   #3
Senior Member

Joined: May 2016
From: USA

Posts: 1,310
Thanks: 551

Quote:
 Originally Posted by ketanco a^2-3a-5=0 then what is the result of: (a^4+25)/(a^2) answer is 19 but can not get there i tried rearranging the first equation and taking square of both sides but it didnt work
There is an alternate avenue of attack. It is more work, but perhaps more intuitive. Use the quadratic formula.

$a^2 - 3a -5 = 0 \implies a = \dfrac{3 \pm \sqrt{3^2 - 4(1)(-\ 5)}}{2 * 1} = \dfrac{3 \pm \sqrt{29}}{2} \implies$

$a^2 = \dfrac{9 \pm 6\sqrt{29} + 29}{4} = \dfrac{38 \pm 6\sqrt{29}}{4} = \dfrac{19 \pm 3\sqrt{29}}{2} \implies$

$a^4 = \dfrac{19^2 \pm 2 * 19 * 3\sqrt{29} + 3^2 * 29}{4} = \dfrac{361 + 261 \pm 114 \sqrt{29}}{4} = \dfrac{311 \pm 57\sqrt{29}}{2}\implies$

$\dfrac{a^4 + 25}{a^2} = \dfrac{\dfrac{311 \pm 57\sqrt{29}}{2} + 25}{\dfrac{19 \pm 3\sqrt{29}}{2}} = \dfrac{\dfrac{361 \pm 57\sqrt{29}}{2}}{\dfrac{19 \pm 3\sqrt{29}}{2}} = \dfrac{361 \pm 57\sqrt{29}}{19 \pm 3\sqrt{29}}.$

If you actually do that division, you get 19 because

$361 \div 19 = 19.$

So you try $57\sqrt{29} \div 19 = 3\sqrt{29}.$

$\therefore \dfrac{361 \pm 57\sqrt{29}}{19 \pm 3\sqrt{29}} = \dfrac{19 \left ( 19 \pm 3\sqrt{29} \right )}{19 \pm 3\sqrt{29}} = 19.$ Tags factorize, fraction 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 Chikis Elementary Math 7 August 31st, 2015 02:31 AM Chikis Algebra 6 October 8th, 2014 06:30 PM Chikis Algebra 3 October 8th, 2014 04:22 PM fantom.1040 Algebra 6 July 16th, 2011 11:31 PM salamatr25 Algebra 12 August 6th, 2010 05:41 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      