Here we observe that coefficient of
x3 is zero.
∴S1=α+β+γ+δ=0⇒α+β=−(γ+δ)=p, say.
And let αβ=q and γδ=r
Hence the given equation equivalent to
(x2−px+q)(x2+px+r)=0
Compare the coefficients of x2,x and constant
q+r−p2=−2,p(q−r)=8,qr=−3
Now using (q+r)2−(q−r)2=4qr⇒(p2−2)2−64p2=−12
⇒(p6−4p4+4p2)+12p2−64=0⇒t3−4t2+16t−64=0 where t=p2
Clearly t=p2=4 satisfies.
∴p=2,q+r=2,q−r=4⇒q=0,r=−1
Hence the given biquadratic is equivalent to.
(x2−2x+3)(x2+2x−1)=0
∴x=−1±√2,−1±√2i