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Question

If α and β are the zeroes of p(x)=4x2+10x+k such that α2+β2+αβ=214, then k =

A
12
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B
4
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C
2
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D
- 12
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Solution

The correct option is B 4
Given, α and β are zeroes of a polynomial.
α+β=104and αβ=k4

Now, it is given that
α2+β2+αβ=214 (α+β)22αβ+αβ=214 (104)2αβ=214 (104)2k4=214 10016k4=214
1004k16=214 1004k=84 10084=4k k=4

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