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Question

If α,β are zeroes of a polynomial 6x2 + x 2, then the polynomial whose zeroes are 2α+3β and 3α+2β , is .

A
6x2+5x1
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B
6x25x+1
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C
6x25x1
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D
6x2+5x+1
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Solution

The correct option is A 6x2+5x1
y=6x2+x2α+β=16 and αβ=26=13Now, Let k1=2α+3βand k2=3α+2βk1+k2=(2α+3β+3α+2β)=5(α+β)=56 andk1×k2=(2α+3β)×(3α+2β)=6(α2+β2)+13αβ=6[(α+β)22αβ]+13αβ=6(α+β)2+αβ=6×136+(13)=1613=16 Required polynomial will be:k[ x2(k1+k2)x+k1k2],where k is non zero real numberk[x2+56x16]k6[6x2+5x1]by putting k = 6, we will get the required answer

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