Let p₁(x)=x^3-2020 x^2+b₁ x+c₁ and p₂(x)=x^3-2021 x^2+b₂ x+c₂ be polynomials having two

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4-2.Quadratic Equations and Inequations

normal

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

$p_1(3)+p_2(1)+4028=0$

$p_1(3)+p_2(1)+4026=0$

$p_1(2)+p_2(1)+4028=0$

$p_1(1)+p_2(2)+4028=0$

(KVPY-2020)

Solution

(a)

Let $p_1(x)=x^3-2020 x^2+b_1 x+c_1$

$=(x-\alpha)(x-\beta)(x-\gamma)$

and $\quad p_2(x)=x^3-2021 x^2+b_2 x+c_2$

$=(x-\alpha)(x-\beta)(x-\delta)$

Since, $p_1(x) \cdot q_1(x)+p_2(x) \cdot q_2(x)$

$=x^2-3 x+2$

On comparing the coefficient of $x^3$, we get $q_1(x)=-q_2(x)=q(x)($ say $)$

So, $(x-\alpha)(x-\beta)[q(x)(\delta-\gamma)]$ $=(x-1)(x-2)$

$\therefore \quad \alpha=1, \beta=2, \gamma=2017$ and $\delta=2018$

$p_1(x)=(x-1)(x-2)(x-2017)$

$\Rightarrow \quad p_1(3)=-4028$

$p_2(x)=(x-1)(x-2)(x-2018)$

$p_2(1)=0$

So,$\quad p_1(3)+p_2(1)+4028=0$

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