BdMO National Higher Secondary 2009/2

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Moon
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BdMO National Higher Secondary 2009/2

Unread post by Moon » Sun Feb 06, 2011 11:32 pm

Problem 2:
Find all integral solutions of the equation: \[\frac{x^2}{2}+\frac{5}{y}=7\]
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Ashfaq Uday
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Re: BdMO National Higher Secondary 2009/2

Unread post by Ashfaq Uday » Wed Oct 19, 2011 12:06 am

(x,y) = (4,-5), (-4,-5), (3,2), (-3,2), (2,-1), (-2,-1)

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samiul_samin
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Re: BdMO National Higher Secondary 2009/2

Unread post by samiul_samin » Tue Feb 20, 2018 11:43 pm

Full Solution
$x^2/2+5/y =7$
$\Rightarrow x^2/2=7y/5$
$\Rightarrow x^2y=14y-10$
$\Rightarrow y(14-x^2)=10$
$10=1×10=2×5=(-2)×(-5)$
$y=1 \Rightarrow 14-x^2=10\Rightarrow x^2=4\Rightarrow x=+5,-5$
$y=2 \Rightarrow 14-x^2=5 \Rightarrow x^2=9 \Rightarrow x=+3,-3$
$y=-5 \Rightarrow 14-x^2=-2 \Rightarrow x^2=16 \Rightarrow x=+4,-4$
So the integral values of ($x,y$) are ($5,1$),($-5,1$),($3,2$),($-3,2$),($4,-5$),($-4,-5$) [Answer]

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samiul_samin
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Re: BdMO National Higher Secondary 2009/2

Unread post by samiul_samin » Wed Mar 06, 2019 11:47 pm

samiul_samin wrote:
Tue Feb 20, 2018 11:43 pm
Full Solution
$\dfrac{x^2}{2}+\dfrac{5}{y} =7$
$\Rightarrow \dfrac{x^2}{2}=7-\dfrac{5}{y}$

$\Rightarrow x^2y=14y-10$

$\Rightarrow y(14-x^2)=10$

$10=1×10=2×5=(-2)×(-5)$

$y=1 \Rightarrow 14-x^2=10\Rightarrow x^2=4\Rightarrow x=+5,-5$

$y=2 \Rightarrow 14-x^2=5 \Rightarrow x^2=9 \Rightarrow x=+3,-3$

$y=-5 \Rightarrow 14-x^2=-2 \Rightarrow x^2=16 \Rightarrow x=+4,-4$

So the integral values of ($x,y$) are ($5,1$),($-5,1$),($3,2$),($-3,2$),($4,-5$),($-4,-5$)
[Answer]

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