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Question
If α,β,γ are the zeros of the polynomial x3+px2+qx+2 such that αβ+1=0, find the value of 2p+q+5
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Solution
Given polynomial = x3+px2+qx+2
The roots of the cubic polynomial are α+β+γ=−p,αβγ=−2,αβ+βγ+γα=q
Also given αβ+1=0⟹αβ=−1
⟹−γ=−2orγ=2[αβγ=−2]
So,α+β+2=−porα+β=−2−pAlso,αβ+βγ+γα=q⟹−1+γ(α+β)=q⟹−1+2(−2−p)=q⟹−1−4−2p=q⟹2p+q+5=0
The roots of the cubic polynomial are α+β+γ=−p,αβγ=−2,αβ+βγ+γα=q
Also given αβ+1=0⟹αβ=−1
⟹−γ=−2orγ=2[αβγ=−2]
So,α+β+2=−porα+β=−2−pAlso,αβ+βγ+γα=q⟹−1+γ(α+β)=q⟹−1+2(−2−p)=q⟹−1−4−2p=q⟹2p+q+5=0
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