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

$\therefore \dfrac{361 \pm 57\sqrt{29}}{19 \pm 3\sqrt{29}} = \dfrac{19 \left ( 19 \pm 3\sqrt{29} \right )}{19 \pm 3\sqrt{29}} = 19.$

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