Given: ax3+bx2+cx+d=0, b2<2ac and a,b,c,d∈R
Let α,β,γ be the roots of ax3+bx2+cx+d=0
∴ sum of roots =α+β+γ=−ba
sum of roots taken two at a time =αβ+βγ+γα=ca
product of roots =α.β.γ=−da
Now, α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)
⇒α2+β2+γ2=(−ba)2−2(ca)
⇒α2+β2+γ2=b2−2aca2
We have given that b2−2ac is negative,
∴α2+β2+γ2<0, which is not possible if all α,β,γ are real. Therefore atleast one roots is non-real, but complex roots occurs in pair. Hence given cubic equation has one real root and two non real roots.