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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 15:42:03 IST
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if a,b,c are the roots of x^3-3x^2+5x-7=0 then 1+ (1/ab) + (1/bc) +(1/ca)=?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 15:52:19 IST
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10/7???
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 15:57:56 IST
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give the method please..
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 16:04:40 IST
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Is it correct??
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 16:08:40 IST
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i dono but its given in the option...so must be correct
but do u mind telling me how u did it???
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 16:15:20 IST
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The expression can be written as (abc+a+b+c)/abc Now,abc is product of the roots which is 7 a+b+c is sum of the roots which is 3 Thus,the given exp. is equal to (7+3)/7 =10/7
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 16:20:34 IST
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i know to find the sum n product only for a second degree eq....
tell me how to find these for a 3rd degree eqn...
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 16:34:26 IST
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ax3 +bx2+cx+d -d/a is product of roots taken 3 at a time c/a is product of roots taken 2 at a time -b/a is sum of roots. I'm not sure about the signs though.
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 16:34:45 IST
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 16:35:23 IST
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Let us consider an equation, ax3 + bx2 + cx + d = 0 Then, a+b+c+d = -b/a ab+bc+ca = c/a abc = -d/a
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 16:45:37 IST
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thanks a lot ...lampard,sboosy,pantpranav...
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