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Algebra Level 4

Let p ( x ) = x 2 − 5 x + a p(x) = x^2-5x +a and q ( x ) = x 2 − 3 x + b q(x) = x^2-3x+ b where a a and b b are positive integers. Suppose gcd ⁡ ( p ( x ) , q ( x ) ) = x − 1 \gcd(p(x), q(x)) = x- 1 and denote k ( x ) = lcm ( p ( x ) , q ( x ) ) k(x) = \text{lcm}(p(x),q(x)) . Find the sum of all distinct roots of ( x − 1 ) + k ( x ) (x-1) + k(x) .


The answer is 4.

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Akshat Sharda
Nov 2, 2015

As the hcf ( p ( x ) , q ( x ) ) = x − 1 ⇒ p ( x ) = x 2 − 5 x + 4 and q ( x ) = x 2 − 3 x + 2 x 2 − 5 x + 4 = ( x − 1 ) ( x − 4 ) and x 2 − 3 x + 2 = ( x − 1 ) ( x − 2 ) k ( x ) = lcm ( p ( x ) , q ( x ) ) = ( x − 1 ) ( x − 2 ) ( x − 4 ) ( x − 1 ) + k ( x ) = ( x − 1 ) + ( x − 1 ) ( x − 2 ) ( x − 4 ) = ( x − 1 ) ( x 2 − 6 x − + 9 ) = ( x − 1 ) ( x − 3 ) 2 Sum of all distinct roots = 1 + 3 = 4 \text{As the hcf}(p(x),q(x))=x-1\Rightarrow p(x)=x^{2}-5x+4 \text{ and }q(x)=x^{2}-3x+2 \\ x^{2}-5x+4=(x-1)(x-4) \text{ and }x^{2}-3x+2=(x-1)(x-2) \\ k(x)=\text{lcm}(p(x),q(x))=(x-1)(x-2)(x-4) \\ (x-1)+k(x)=(x-1)+(x-1)(x-2)(x-4)=(x-1)(x^{2}-6x-+9)=(x-1)(x-3)^{2} \\ \text{Sum of all distinct roots}=1+3=\boxed{4}

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