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Question

Solve: x42x2+8x3=0 and find sum of its real roots ?

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Solution

Here we observe that coefficient of x3 is zero.
S1=α+β+γ+δ=0α+β=(γ+δ)=p, say.
And let αβ=q and γδ=r
Hence the given equation equivalent to
(x2px+q)(x2+px+r)=0
Compare the coefficients of x2,x and constant
q+rp2=2,p(qr)=8,qr=3
Now using (q+r)2(qr)2=4qr(p22)264p2=12
(p64p4+4p2)+12p264=0t34t2+16t64=0 where t=p2
Clearly t=p2=4 satisfies.
p=2,q+r=2,qr=4q=0,r=1
Hence the given biquadratic is equivalent to.
(x22x+3)(x2+2x1)=0
x=1±2,1±2i

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