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Suppose we are given that $a+b+c = A $, $ab+bc+ca = B$ , $abc = C$ , We can observe that if we solve by eliminating two variables out of three(a,b,c) from three equations we would end up getting $x^3 - Ax^2 +Bx -C =0$ a cubic equation which is satisfied by the last one not eliminated,

  • So if we do the same with a cubic equation with two roots equal and going backwards we would have two variables and three equations so in general for a cubic equation whose roots we know are supposed to be having a double root can be easily solved as compared to the ones which are distinct in general ?
  • So without cardano reduction stuff we can solve for all roots in this special equal roots case (double) easily ?
Paracetamol
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  • Are you assuming the coefficients of your cubic are integers? Or maybe rationals, which is reducible to being integers? – coffeemath Jun 28 '22 at 19:23
  • It can be for any polynomial x^3 +... Which has real coefficients i meant ? @coffeemath whose root nature we alrwady know from before – Paracetamol Jun 28 '22 at 19:25
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    @Paracetamol The cubic has a double root iff $,\gcd(P,P'),$ has degree $,\ge 1,$, so in that case you have to solve an equation of degree at most $,2,$. – dxiv Jun 28 '22 at 19:26
  • @dxiv yeah that i know i was not referring to that , i mean if i know already from start that the cubic has two roots equal , then if we are told to solve the cubic , will we need to use cardano to get reduce cubic etc .. or just solving the system of equations will lead the answer for both roots . Since we have three equations twl variables unlike the case of all three distinct where we get back to original equation while solving – Paracetamol Jun 28 '22 at 19:28
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    @Paracetamol You don't have to solve the system. You can either calculate the $,\gcd,$ which involves just Euclidean division, or directly solve $,P'(x)=0,$ and verify which of the two roots is also a root of $,P(x),$. – dxiv Jun 28 '22 at 19:31
  • Oh yeah right , understood nice method thanks – Paracetamol Jun 28 '22 at 19:33

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