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Question

If α,β are roots of the equation ax2+bx+c=0, then the quadratic equation whose roots are 1(aα+b)2,1(aβ+b)2, is

A
(c2)x2+(2acb2)x+1=0
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B
(a2c2)x2+(2acb2)x+1=0
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C
(a2)x2+(2acb2)x+1=0
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D
(a2c2)x2+(2ac+b2)x+1=0
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Solution

The correct option is B (a2c2)x2+(2acb2)x+1=0
We know that α,β are the roots of
ax2+bx+c=0 (1)
Now, aα2+bα+c=0
α(aα+b)+c=0
(aα+b)=cα, (aβ+b)=cβ
Therefore, 1(aα+b)2=α2c2
and 1(aβ+b)2=β2c2

Let y=x2c2, where x=α,β
c2y=x2x=cy
Putting this in equation (1),
a(cy)2+b(cy)+c=0ac2y+bcy+c=0(ac2y+c)2=(bcy)2a2c4y2+(2ac3b2c2)y+c2=0(a2c2)y2+(2acb2)y+1=0

Hence, the required quadratic equation is,
(a2c2)x2+(2acb2)x+1=0

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