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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: quadratic..
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mathwiz (46)

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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)=?
    
LAMPARD (1142)

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10/7???

MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD...
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mathwiz (46)

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give the method please..
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LAMPARD (1142)

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Is it correct??

MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD...
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mathwiz (46)

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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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LAMPARD (1142)

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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

MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD...
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mathwiz (46)

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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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LAMPARD (1142)

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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.

MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD...
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sboosy (3009)

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ax^3+bx^2+cx+d=0 \\ \\ \mbox{Let the roots of the equation be} \ \alpha,\beta,\gamma \\ \\ \mbox{Sum of the roots} = \frac{\mbox{-coeff.of} \ x^2}{\mbox{coeff.of} \ x^3} \Rightarrow \alpha+\beta+\gamma = \frac{-b}{a} \\ \\ \mbox{Sum taken two at a time} =  \frac{\mbox{coeff.of} \ x}{\mbox{coeff.of} \ x^3} \ \Rightarrow \alpha\beta+\beta\gamma+\gamma\alpha = \frac{c}{a} \\ \\ \mbox{Product of the roots} \ = \frac{-\mbox{Constant term}}{\mbox{coeff.of} \ x^3} \ \Rightarrow \alpha\beta\gamma = \frac{-d}{a} \\
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pantpranav (341)

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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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mathwiz (46)

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thanks a lot ...lampard,sboosy,pantpranav...
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