Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.

Roots of Equation and Vieta’s formula – AIME I, 1996

Suppose that the roots of \(x^{3}+3x^{2}+4x-11=0\) are a,b and c and that the roots of \(x^{3}+rx^{2}+sx+t=0\) are a+b,b+c and c+a, find t.

  • is 107
  • is 23
  • is 840
  • cannot be determined from the given information

Key Concepts


Roots of Equation

Vieta s formula

Check the Answer

But try the problem first…

Answer: is 23.

Suggested Reading

AIME I, 1996, Question 5

Polynomials by Barbeau

Try with Hints

First hint

With Vieta s formula


\(\Rightarrow a+b+c=-3\), \(ab+bc+ca=4\) and \(abc=11\)

Second Hint

Let a+b+c=-3=p

here t=-(a+b)(b+c)(c+a)

\(\Rightarrow t=-(p-c)(p-a)(p-b)\)

\(\Rightarrow t=-f(p)=-f(-3)\)

Final Step



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