The correct option is C (a2c2)x2+(2ac−b2)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=α2c21(aβ+b)2=β2c2
The quadratic equation whose roots are α2c2,β2c2 is
Let y=x2c2, where x=α,β
c2y=x2⇒x=c√y
Putting this in the equation (1),
a(c√y)2+b(c√y)+c=0⇒ac2y+bc√y+c=0⇒(ac2y+c)2=(−bc√y)2⇒a2c4y2+(2ac3−b2c2)y+c2=0⇒(a2c2)y2+(2ac−b2)y+1=0
Hence, the required quadratic equation is,
(a2c2)x2+(2ac−b2)x+1=0