Given, equation is x4−2x3+x2+4x−6=0 Let the roots of this equation be α,β,γ and δ. Sum of the roots =α+β+γ+δ=‌
−(−2)
1
=2 Sum of the products of roots ‌=αβ+αγ+αδ+βγ+βδ+γδ ‌=‌
1
1
=1 Sum of products of roots taken three at a time ‌=αβγ+αβδ+αγδ+βγδ ‌=‌
−4
1
=−4 Product of roots =αβγδ=‌
−6
1
=−6 Since, the sum of two roots be zero, i.e., ‌α+β‌=0 ⇒‌β‌=−α So, γ+δ=2‌‌ [from Eq. (i)] And, ‌α(−α)+αγ+αδ+(−αγ)+(−α)δ+γδ=1 ‌⇒‌‌−α2+αγ+αδ−αγ−αδ+γδ=1 ‌⇒‌‌−α2+γδ=1‌‌...(ii) And ‌‌α(−α)γδ=−6 ⇒‌‌−α2γδ=−6 From Eqs. (ii) and (iii), we get ‌γδ=1+α2 ⇒α2(1+α2)=6 ⇒α4+α2=6 Let y=α2, then we get ‌y2+y−6=0 ⇒(y+3)(y−2)=0 ⇒y=−3‌ and ‌y=2 Since, α2 must be non-negative, ‌y=α2=2 ‌‌ Then, ‌γδ=1+α2=1+2=3 ‌‌ Now, ‌(γ+δ)2=γ2+δ2+2γδ ‌⇒‌‌γ2+δ2=(γ+δ)2−2γδ ‌‌‌=22−2(3)=4−6=−2