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Question

If α and β are the roots of the equation ax2+bx+c=0 and if px2+qx+r=0 has roots 1αα and 1ββ, then r is

A
a+2b
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B
a+b+c
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C
ab+bc+ca
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D
abc
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Solution

The correct option is B a+b+c
The equation with roots 1α and 1β
=a(1x)2+b(1x)+c
=cx2+bx+a=0 ..(1)
Now 1αα=1α1
Similarly 1ββ=1β1
Therefore the quadratic equation containing these roots is
c(x+1)2+b(x+1)+a
=cx2+(b+2c)x+a+b+c=0
By comparing coefficients we get with px2+qx+r=0 we get
r=a+b+c

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