The number of integral solutions of the equation 7(y+(1/y))−2(y^2 +(1/y^2 ))=9 are?


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Question Number 10195 by priyank last updated on 30/Jan/17

$T h e n u m b e r o f i n t e g r a l s o l u t i o n s o f t h e e q u a t i o n$ $7 (y + \frac{1}{y}) - 2 (y^{2} + \frac{1}{y^{2}}) = 9 are ?$
Commented by prakash jain last updated on 30/Jan/17

$y + \frac{1}{y} = u$ $(y^{2} + \frac{1}{y^{2}}) = y^{2} + \frac{1}{y^{2}} + 2 - 2 = {(y + \frac{1}{y})}^{2} - 2 = u^{2} - 2$ $7 u - 2 (u^{2} - 2) = 9$ $7 u - 2 u^{2} + 4 = 9$ $2 u^{2} - 7 u + 5 = 0$ $(2 u - 5) (u - 1) = 0$ $u = \frac{5}{2} o r u = 1$ $y + \frac{1}{y} = \frac{5}{2} \Rightarrow 2 y^{2} + 2 = 5 y$ $2 y^{2} - 5 y + 2 = 0 ⇎ (2 y - 1) (y - 2) = 0$ $y = 2 o r y = \frac{1}{2}$ $y + \frac{1}{y} = 1 \Rightarrow y^{2} - y + 1 = 0 \Rightarrow y = \frac{1 \pm \sqrt{- 3}}{2}$ $o n l y o n e i n t e g e r s o l u t i o n y = 2$
	Answered by FilupSmith last updated on 30/Jan/17

	$Do you mean integer solution ?$ $7 y + \frac{7}{y} - 2 y^{2} - \frac{2}{y^{2}} = 9$ $7 y^{3} + 7 y - 2 y^{4} - 2 = 9 y^{2}$ $- 2 y^{4} + 7 y^{3} - 9 y^{2} + 7 y - 2 = 0$ $2 y^{4} - 7 y^{3} + 9 y^{2} - 7 y + 2 = 0$ $(2 y - 1) (y - 2) (y^{2} - y + 1) = 0 (from wolfram alpha)$ $(1) 2 y = 1 \Rightarrow y = \frac{1}{2}$ $(2) y = 2$ $(3) y = \frac{1 \pm i \sqrt{3}}{2}$ $∴ y has 1 integer root$
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