Jess-06-hisec-Q.10

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rakeen
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Jess-06-hisec-Q.10

Unread post by rakeen » Sat Dec 18, 2010 11:05 am

$m$ er maan koto hole $x^2 + xy +5x + m + 5$ ke mulod utpadoke bisleshon kora jabe?





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Re: Jess-06-hisec-Q.10

Unread post by Moon » Sun Jan 09, 2011 10:13 pm

লোকজন কি জান রাকিন কই গেল? ডিভিশনাল নিয়া আপসেট নাকি? যাইহোক পোস্টের জন্য হিন্ট:
$x^2 + xy +5x + m + 5=(ax+by+c)(dx+e)$. Compare the co efficients. :)
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Re: Jess-06-hisec-Q.10

Unread post by amlansaha » Tue Dec 06, 2011 2:20 am

let $x^2+xy+5x+m+5=(ax+by+c)(dx+e)=adx^2+bdxy+(cd+ae)x+bey+ce$
comparing the co efficiantes we get
$ad=1, bd=1, cd+ae=5, be=0, ce=m+5$
now, as $b\neq 0$ [bd=1] $e=0$. so $m+5=0$
so, $m=-5$ :D :D :D
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