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if the root of the equation px2+qx+r=0 are real and of the form x/x+1 and x+1/x, then show that (p+q+r/2=4p2+_q2+4pq​

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Asked on December 20, 2019 by

Prasanna Bargali

If the roots of the equation px

2

+qx+r=0, where 2p,q,2r are in G.P are of the α

2

,4α−4 than the value of 2p+4q+7r is :

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ANSWER

Given 2p,q,2r are in GP⟹q

2

=4pr

α,4α−4 are the roots of the equation px

2

+qx+r=0

As we know that

Difference between the roots of ax

2

+bx+c=0 is

a

b

2

−4ac

⟹α

2

−(4α−4)=

p

q

2

−4pr

=

p

4pr−4pr

=0

⟹α

2

−4α+4=0⟹(α−2)

2

=0⟹α=2

The roots are 4,4So the equation is x

2

−8x+16=0

So p=1,q=−8,r=16

2p+4q+7r=2−32+112=82