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.

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

Answer: is 23.

AIME I, 1996, Question 5

Polynomials by Barbeau

Try with Hints

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 \(abc=11\)

Let a+b+c=-3=p

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

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

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

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

=23.

Quadratic Equation | SMO, 2012 | Junior Section

Try this beautiful problem from Singapore Math Olympiad, 2012, Junior Senior based on Quadratic Equation.

Quadratic Equation - Singapore Mathematics Olympiad, 2012

Consider the equation

\(\sqrt {3x^2 - 8x + 1} + \sqrt {9x^2 - 24x - 8}\) = 3.

Key Concepts

Quadratic Function

Analysis of Number

Root of Equation

Check the Answer

Answer: 9

Singapore Mathematics Olympiad

Challenges and Thrills - Pre - College Mathematics

Try with Hints

As the first hint we can assume :

y =\(3x^2 - 8x + 1 \) then the equation becomes

y + \(\sqrt {3y^2 -11}\) = 3.

Lets try to do the rest of the sum ....................

If we are still stuck after the first hint we can say :

Then \( \sqrt {3y^2 - 11}\) = 3 - y.

Lets square the both sides , we have \( 3 y^2 - 11 = 9 -6y + y^2 \) ,

Then y = 2 or y = -5

Now solve \( 3x^2 - 8x +1 = 2^2 \)

Then x =3 and \(x = - 3^{-1}\)

Hence k = \(\frac {3}{3^{-1}}\) = 9 (Answer )

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