Question

Let p₁(x) = x³ - 2020x² + b₂x+c₁ and p₂(x) = x³ - 2021x² + b₂x + c₂ be polynomials having two common roots a and B. Suppose there exist polynomials q₁(x) and 92(x) such that p₁(x)q₁(x) + p₂(x)q₂(x) = x²-3x + 2. Then the correct identity is

Answer 1

Answer:

The equations x3+5x2+px+q=0and x3+7x2+px+r=0 have 2 common roots, then find the third root of both equations

From the first equation we can say, αβ+βγ+γα=p/1=p. Similarly from the second equation we know, αβ+βδ+δα=p/1=p

Hence,

αβ+βδ+δα=αβ+βγ+γα

δ(β+α)=γ(β+α)

δ=γ

Hence the third root of both equations should be equal, but α+β+γ=−5 and α+β+δ=−7. Now, where did I go wrong?

Answer 2

The correct identity is $p_{1} (3)+p_{2} (1)+4028=0$.

Given:

$p_{1} (x)=x^{3} -2020x^{2} +b_{1} x+c_{1}$

$p_{2} (x)=x^{3} -2021x^{2} +b_{2} x+c_{2}$

$p_{1}(x) q_{1}(x) +p_{2}(x) q_{2}(x) =x^{2} -3x+2$

To Find:

The correct identity

Solution:

We can simply solve this problem by using the following mathematical process.

As

$p_{1}(x) q_{1}(x) +p_{2}(x) q_{2}(x) =x^{2} -3x+2$

Also

$p_{1}(x) -p_{2}(x) =x^{2} +x(b_{1} -b_{2} )+(c_{1} -c_{2} )$

Therefore,

$q_{1} (x)=1$ and $q_{2} (x)=-1$

Now,

$p_{1} (x)=x^{3} -2020x^{2} +b_{1} x+c_{1}$

Since the polynomial is of degree = 3, so there will be three roots.

Let the roots be 1,2 and t

So,

t+3=2020

t=2017

Therefore,

$p_{1} (x)=(x-1)(x-2)(x-2017)$

Similarly

$p_{2} (x)=(x-1)(x-2)(x-2018)$

Putting the values 3 and 1 in $p_{1}$ and $p_{2}$ respectively we have,

$p_{1} (3)+p_{2} (1)+4028=0$

$p_{1} (3)=-4028$

$p_{2} (1)=0$

So, the equation is satisfied.

Hence, the correct identity is $p_{1} (3)+p_{2} (1)+4028=0$.

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