Categories
AIME I Algebra Arithmetic Functional Equations Math Olympiad USA Math Olympiad

Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5

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

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


Functions

Roots of Equation

Vieta s formula

Check the Answer


But try the problem first…

Answer: is 23.

Source
Suggested Reading

AIME I, 1996, Question 5

Polynomials by Barbeau

Try with Hints


First hint

With Vieta s formula

\(f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0\)

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

\(t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]\)

=23.

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.