If α,β,γ,σ are the roots of the equation x4+4x3−6x2+7x−9=0 then the value of 1+α21+β21+γ21+σ2 is

Since α,β,γ,σ are the roots of the given equation ∴ x4+4x3−6x2+7x−9=(x−α)(x−β)(x−γ)(x−σ)Putting x = i and then x == -i, we getand 1−4i+6+7i−9=(i−α)(i−β)(i−γ)(i−σ)1+4i+6−7i−9=(−i−α)(−i−β)(−i−γ)(−i−σ)Multiplying these two, we get(−2+3i)(−2−3i)=1+α21+β21+γ21+σ2⇒ 13=1+α21+β21+γ21+σ2