Categories
AMC 10 Math Olympiad USA Math Olympiad

Roots of cubic equation | AMC-10A, 2010 | Problem 21

Try this beautiful problem from Algebra:Roots of cubic equation from AMC-10A, 2010. You may use sequential hints to solve the problem

Try this beautiful problem from Algebra based Roots of the cubic equation.

Roots of cubic equation – AMC-10A, 2010- Problem 21


The polynomial \(x^3-ax^2+bx-2010\) has three positive integer roots. What is the smallest possible value of \(a\)?

  • \(98\)
  • \(78\)
  • \(83\)
  • \(76\)
  • \(90\)

Key Concepts


Algebra

Vieta’s Relation

roots of the equation

Check the Answer


Answer: \(78\)

AMC-10A (2010) Problem 19

Pre College Mathematics

Try with Hints


The given equation is \(x^3-ax^2+bx-2010\).we have to find out the smallest possible value of \(a\).From Vieta’s Relation we know that if \(r_1,r_2,r_3\) are the roots of equation \(ax^3+bx^2+cx+d=0\) then \(r_1 +r_2+r_3= -\frac{b}{a}\) and \(r_1 r_2 r_3=-\frac{d}{a}\)

can you finish the problem……..

Therefore \(a\) is the sum of the three roots of the polynomial \(x^3-ax^2+bx-2010\). and \(2010\)  is the product of the three integer roots.Now the factors of \(2010\) =\(2 \times 3 \times 5 \times 67\).there are only three roots to the polynomial so out of four roots we have to choose three roots such that two of the four prime factors must be multiplied so that we are left with three roots. To minimize \(a\), \(2\) and \(3\) should be multiplied,

can you finish the problem……..

Therefore the value of \(a\)=\(6+5+67=78\)

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *