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Question

If α and β are the zeroes of the quadratic polynomial 2S28S+4, then the value of αβ+βα+2(1α+1β)+3αβ is ___.

A
8
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B
4
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C
16
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D
12
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Solution

The correct option is C 16
Given, f(S)=2S28S+4.
Let α and β be its zeroes.
α+β=82=4and αβ=42=2
Now,
αβ+βα+2(1α+1β)+3αβ

=α2+β2αβ+2(α+βαβ)+3αβ

=(α+β)22αβαβ+2(α+βαβ)+3αβ

[(α+β)2=α2+β2+2αβ]

=(4)22×22+2(42)+3×2

= 6 + 4 + 6 = 16

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