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

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

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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

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