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November 13th, 2018, 12:09 PM   #1
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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
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November 13th, 2018, 12:51 PM   #2
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You were on the right track, but did not go far enough.

$a^2 - 3a - 5 = 0 \implies a^2 = 3a + 5 \implies$

$a^4 = 9a^2 + 30a + 25 = 9(3a + 5) + 30a + 25 = 57a + 70 \implies$

$a^4 + 25 = 57a + 95 = 19(3a + 5) \implies \dfrac{a^4 + 25}{a^2} = \dfrac{19(3a + 5)}{3a + 5} = 19.$
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November 13th, 2018, 02:37 PM   #3
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Quote:
Originally Posted by ketanco View Post
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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