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Question

If the difference between the roots of x2+2px+q=0 is two times the difference between the roots of x2+qx+p4=0, where pq, then

A
pq+1=0
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B
pq1=0
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C
p+q1=0
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D
p+q+1=0
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E
q4p+1=0
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Solution

The correct option is D p+q+1=0
Let the roots of the equation x2+2px+q=0 be α1 and β1 while roots of the equation x2+qx+p4=0 are α2 and β2.

Then, α1+β1=2p

and α1β1=q

and α2+β2=q

and α2β2=p4

Now, (α1β1)2=(α1+β1)24α1β1

=4p24q=4(p2q)

α1β1=2p2q

Also, (α2β2)2=(α2+β2)24α2β2

=q24×p4=q2p

α2β2=q2p

It is given that, (α1β1)=2(α2β2)

2p2q=2q2p

p2q=q2p

p2q=q2p

p2q2+pq=0

(pq)(p+q)+(pq)=0

(pq)(p+q+1)=0

pq=0 or p+q+1=0

but pq0 as pq

Therefore, p+q+1=0

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