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November 13th, 2018, 11:09 AM  #1 
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  factorize fraction
a^23a5=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: 550 
You were on the right track, but did not go far enough. $a^2  3a  5 = 0 \implies a^2 = 3a + 5 \implies$ $a^4 = 9a^2 + 30a + 25 = 9(3a + 5) + 30a + 25 = 57a + 70 \implies$ $a^4 + 25 = 57a + 95 = 19(3a + 5) \implies \dfrac{a^4 + 25}{a^2} = \dfrac{19(3a + 5)}{3a + 5} = 19.$ 
November 13th, 2018, 01:37 PM  #3  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
$a^2  3a 5 = 0 \implies a = \dfrac{3 \pm \sqrt{3^2  4(1)(\ 5)}}{2 * 1} = \dfrac{3 \pm \sqrt{29}}{2} \implies$ $a^2 = \dfrac{9 \pm 6\sqrt{29} + 29}{4} = \dfrac{38 \pm 6\sqrt{29}}{4} = \dfrac{19 \pm 3\sqrt{29}}{2} \implies$ $a^4 = \dfrac{19^2 \pm 2 * 19 * 3\sqrt{29} + 3^2 * 29}{4} = \dfrac{361 + 261 \pm 114 \sqrt{29}}{4} = \dfrac{311 \pm 57\sqrt{29}}{2}\implies$ $\dfrac{a^4 + 25}{a^2} = \dfrac{\dfrac{311 \pm 57\sqrt{29}}{2} + 25}{\dfrac{19 \pm 3\sqrt{29}}{2}} = \dfrac{\dfrac{361 \pm 57\sqrt{29}}{2}}{\dfrac{19 \pm 3\sqrt{29}}{2}} = \dfrac{361 \pm 57\sqrt{29}}{19 \pm 3\sqrt{29}}.$ 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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