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Question

Let α1,α2 and β1,β2 be the roots of ax2+bx+c=0 and px2+qx+r=0 respectively. If the system of equations α1y+α2z=0 and β1y+β2z=0 has a non-trivial solution, then which of the following options is CORRECT ?
  1. abc=pqr
  2. a2pr=q2bc
  3. b2pr=q2ac
  4. a2qr=p2bc


Solution

The correct option is C b2pr=q2ac
ax2+bx+c=0
α1+α2=ba, α1α2=ca
and px2+qx+r=0
β1+β2=qp, β1β2=rp

α1y+α2z=0, β1y+β2z=0 have a non-trivial solution.
α1α2β1β2=0

α1β2α2β1=0
α1β1=α2β2=k (say)

Now, α1+α2=ba
k(β1+β2)=ba
k(qp)=ba
k=pbqa     (1)

α1α2=ca
k2β1β2=ca 
k2(rp)=ca
k2=pcar     (2)

From (1) and (2),
p2b2q2a2=pcar
b2pr=q2ac 

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