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Question:
If α and β are the zeroes of the quadratic polynomial p(s) = 3s2 – 6s 4, find the value of ∝ 𝛽 𝛽 𝛼 2( 1 𝛼 1 𝛽 ) 3𝛼𝛽.
Answer:

Given, α and β are the zeroes of the quadratic polynomial p(s) = 3s2 – 6s + 4

Now, α + β = -(-6)/3

=> α + β = 6/3

=> α + β = 2

and α * β = 4/3

Now, (α/β) + (β/α) + 2(1/α + 1/β) + 3 * α * β

= (α2 + β2 )/α * β + 2(α + β)/α * β + 3 * α * β

= {(α + β)2 - 2α * β}/α * β + 2(α + β)/α * β + 3 * α * β

= {22 - 2 *4/3 }/(4/3) + (2 * 2)/(4/3) + 3 * (4/3)

= (4 - 8/3)/(4/3) + 4/(4/3) + 4

= (4/3)/(4/3) + (4*3)/4 + 4

= 1 + 3 + 4

= 8

So, (α/β) + (β/α) + 2(1/α + 1/β) + 3 * α * β = 8

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