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in Quadratic Equations by (15 points)
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Let \( p_{1}(x)=x^{3}-2020 x^{2}+b_{1} x+c_{1} \) and \( p_{2}(x)=x^{3}- \) \( 2021 x^{2}+b_{2} x+c_{2} \) be polynomials having two common roots \( \alpha \) and \( \beta \). Suppose there exist polynomials \( q_{1}(x) \) and \( q_{2}(x) \) such that \( p_{1}(x) q_{1}(x)+p_{2}(x) q_{2}(x)=x^{2}-3 x+2 \). Then the correct identity is

(A) \( p_{1}(3)+p_{2}(1)+4028=0 \) 

(B) \( p_{1}(3)+p_{2}(1)+4026=0 \) 

(C) \( P_{1}(2)+P_{2}(1)+4028=0 \) 

(D) \( p_{1}(1)+p_{2}(2)+4028=0 \)

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1 Answer

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Let α, ß, \(\gamma\) are roots of P1(x) and α, ß, \(\delta\) are roots of P2(x).

∴ α + ß + \(\gamma\) = 2020

and α + ß + \(\delta\) = 2021

⇒ \(\delta-\gamma\) = 1

Also αß + ß\(\gamma\) + \(\gamma\)α = b1

αß + ß\(\delta\) + \(\delta\)α = b2

⇒ ß(\(\delta\) - \(\gamma\)) + α(\(\delta\) - \(\gamma\)) = b2 - b1

⇒ ß + α = b2 - b1

Also, αß\(\gamma\) = -C1

αß\(\gamma\) = -C2

⇒ αß = C1 - C2

Given that p1(x) q1(x)+ p2(x)q2(x) = x2 - 3x + 2

⇒ (x - α) (x - ß) (x - γ)q1 (x) + (x - α) (x - ß) (x - δ )q2(x) = (x - 1) (x - 2)

⇒ (x - α) (x - ß) ((x - γ )q1(x) + (x - δ)q2(x)) = (x - 1) (x - 2)

∃ q1(x), q2(x) such that (x - γ)q1(x) + (x - δ) q2(x) = 1

α = 1, ß = 2

Therefore, r = 2017

δ = 2018

∴ p1(x) = (x- 1) (x- 2) (x - 2017)

p2(x) = (x - 1)(x - 2) (x - 2018)

p1(3) = 2 x 1 x 2014 = -4032

p2(1) = 0

p1(3) + p2(1) = -4032

⇒ p1(3) + p2(1) + 4032 = 0

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