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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Jun 2008 18:54:46 IST
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if da q eq (b2 + c2)x2 -2(a+b)x + (c2 + a2) =0 has equal roots ,then
a) a,b,c are in GP b)a,c,b are in GP c) a,c,b are in AP d) a,b,c are in AP
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DONT AIM FOR PERFECTION JUST ACHIEVE IT |
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13 Jun 2008 19:09:46 IST
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plzzz reply
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DONT AIM FOR PERFECTION JUST ACHIEVE IT |
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abc r in A.P.
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if u live to be a hundred
i want to live to be hundred minus 1 day
so i never have to live without u
.
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13 Jun 2008 19:43:03 IST
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naaa da ans is acb r in GP plzzz help
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DONT AIM FOR PERFECTION JUST ACHIEVE IT |
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14 Jun 2008 13:12:39 IST
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re:if da q eq (b2 + c2)x2 -2(a+b)x + (c2 + a2) =0 has equal roots ,then
a) a,b,c are in GP b)a,c,b are in GP c) a,c,b are in AP d) a,b,c are in AP
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DONT AIM FOR PERFECTION JUST ACHIEVE IT |
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14 Jun 2008 17:24:04 IST
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Ah..I tried using the fact that discriminant is 0 but didn't get anything out of it as in the options.But I don't think a,c,b are in G.P is the right answer bcoz if it is so,then c2=ab.
Hence,the eqn. becomes b(a+b)x2-2(a+b)x+a(a+b)=0
 (a+b)=0 or bx2-2x+a=0
Here,a+b is non zero bcoz in that case it results that (b2+c2)(x2+1)=0 which has no real roots.It results that a=b=c=0,which is not the case.
So,bx2-2x+a=0 shud have equal roots which is true when ab=1.
Thus,It satisfies only some cases in which ab=c2=1 but not all.
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MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC. |
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