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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Nov 2007 22:39:24 IST
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How do you solve cubic equation of the form: ax3 + bx2 + cx + d = 0 is there any formula like there is for quadratic??
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Nov 2007 22:46:22 IST
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no there is no exact formula but u can solve it this way. if p,q, and r are the roots then p+q+r=-b/a pq+qp+rp=c/a pqr=-d/a solving the above three equations u can get the roots rate if useful
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Nov 2007 22:49:22 IST
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but then if u solve it usin this, u will again get a cubic eqn...in p q or r..!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Nov 2007 22:50:32 IST
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let x1,x2,x3 be its roots then, x1+x2+x3 = -b/a x1x2+ x2x3 + x3x1 = c/a x1x2x3 = -d/a
this is only we know abt equation................
rate if satisfied.............otherwise nudge me.
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LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.
Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
Albus Dumbledore
Harry Potter and the Prisoner of Azkaban movie
There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.
-- Albert Einstein
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Dec 2007 20:53:51 IST
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yes. there is a formula first u put x=y-b/3a then substitute in the cubic equation then you get the equation which do not have x2 term
that is a cubic in x with co-efficient of x2=0 then if u get a cubic like this, y3+ay=b,put b=s3-t3&a=3st(assumption) then,u observe that,y=s-t satisfies y3+ay=b hence s=a/3t solve for t then solve for s hence u can get y from that,u can get x hence cubic equation is solved please rate me
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Dec 2007 20:55:45 IST
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this is the exact formula for solving cubic & quatic equations also rate me
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