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How Do I solve this eqns?

$$x+y+z = A$$ $$xyz = B$$ $$x^2+y^2+z^2 = C$$

I have tried it in this way,,,

$$yz = B/x = P$$ $$y+z = A-x = Q$$ $$y(Q-y) = P$$ $$\implies y^2-Qy+p = 0$$

I can't figure out what to do next., Solving the this quadratic equation for y and z and putting the values in 3, doesnt seem to work. cause there will be sqrt in the eqn.

Tamim Addari
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2 Answers2

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We have $$xy+yz+zx=\frac{1}{2}((x+y+z)^2-x^2-y^2-z^2)=\frac{1}{2}(A^2-C)=C'$$ so by Vieta's formulas $x,y$ and $z$ are solutions of the polynomial equation: $$t^3-At^2+C't-B=0$$

  • :( Is there any way to better the guess of one root in cubic equation? – Tamim Addari Aug 16 '13 at 06:03
  • In the case where the coefficients $A,B$ and $C$ are integers we can use the rational root theorem to find a possible rational root. –  Aug 16 '13 at 06:37
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Hint: $xy + yz + zx = (x+y+z)^2 - (x^2 + y^2 + z^2) = \frac{A^2 - C}{2}$

Consider the equation $M^3 - AM^2 + ( \frac{A^2-C} {2} )M - B=0$. It has roots $x, y$ and $ z$.

Calvin Lin
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