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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
If α,β are zeroes of a polynomial 6x2 + x – 2, then the polynomial whose zeroes are 2α+3β and 3α+2β , is
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Solution
The correct option is A 6x2+5x−1
y=6x2+x−2⇒α+β=−16 and αβ=−26=−13Now, Let k1=2α+3βand k2=3α+2β∴k1+k2=(2α+3β+3α+2β)=5(α+β)=−56 andk1×k2=(2α+3β)×(3α+2β)=6(α2+β2)+13αβ=6[(α+β)2−2αβ]+13αβ=6(α+β)2+αβ=6×136+(−13)=16−13=−16∴ Required polynomial will be:k[ x2−(k1+k2)x+k1k2],where k is non zero real number⇒k[x2+56x−16]⇒k6[6x2+5x−1]by putting k = 6, we will get the required answer
y=6x2+x−2⇒α+β=−16 and αβ=−26=−13Now, Let k1=2α+3βand k2=3α+2β∴k1+k2=(2α+3β+3α+2β)=5(α+β)=−56 andk1×k2=(2α+3β)×(3α+2β)=6(α2+β2)+13αβ=6[(α+β)2−2αβ]+13αβ=6(α+β)2+αβ=6×136+(−13)=16−13=−16∴ Required polynomial will be:k[ x2−(k1+k2)x+k1k2],where k is non zero real number⇒k[x2+56x−16]⇒k6[6x2+5x−1]by putting k = 6, we will get the required answer