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Given the equation:

$$x^{3}+ mx^{2}+ n= 0\left ( * \right )$$

Find $m,\,n$ so the equation $\left ( * \right )$ has 3 distinct non-zero real roots $a,\,b,\,c$ satisfying $$\frac{a^{4}}{a^{3}- 2\,n}+ \frac{b^{4}}{b^{3}- 2\,n}+ \frac{c^{4}}{c^{3}- 2\,n}= 3$$

I try to use the Carnado's method and Wolfram Alpha, I found:

$$a\approx -1,\,4,\,b\approx 2,\,8,\,c\approx 2,\,8,\,m\approx-4,\,2,\,n\approx-10,\,976$$

Thus, we can assume $- 2\,a= b= c$, and replace on the equation. But the expression is too large and I can't continue! Help me! Thanks!

  • Where does problem come from? – quasi Jun 18 '18 at 04:22
  • This may help: using Mathematica i found the roots of the equation and substituted them in the imposed condition together with simplify command, I get $−m$ as an answer which means that $m=−3$, so the question becomes what are the values of $n$ such that the equation has 3 distinct non-zero real roots. –  Jun 18 '18 at 05:11
  • @dang dang haj: Are you sure you stated the problem correctly? As stated, there are infinitely many pairs $(m,n)$ of real numbers for which the specified conditions hold. – quasi Jun 18 '18 at 05:16

1 Answers1

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You should start writing down Vieta's Formulas, and use these constantly :

$$a+b+c=-m,\quad ab+bc+ca=0,\qquad abc=-n$$

If you now look at a single denominator, you can rewrite it as :

$$ a^3-2n=a^3+2abc = a(a^2+2bc)$$ $$\qquad=a\left(a^2-2(ab+ac)\right)=a^2(3a+2m)$$

So, the complete fraction becomes now :

$$\frac{a^2}{3a+2m}+\frac{b^2}{3b+2m}+\frac{c^2}{3c+2m}=3$$

If you place everything on a single denominator, and you juggle around with Vieta's formulas, you should get something in the following line :

$$-\frac{4 m^4 + 9 m n - 6 m^2 n}{4 m^3 + 27 n}=-\frac{4 m^3 + (9 - 6 m) n}{4 m^3 + 27 n}\cdot m=3$$

Although this has many solutions, a possible integer solution could be obtained by making this division work, i.e: $m=-3$ and any integer $n$ different from zero.

Eg. $x^3-3x^2+2=0$, has zero's $1,1-\sqrt{3},1+\sqrt{3}$ which satisfy this condition.

Furthermore, since you requested only to have real roots, your cubic discriminant should be bigger than zero. In this case, you have $\Delta=(4m^3+27n)n>0$

kvantour
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