Let f(x)=ax2+bx+c=0 has roots α,β
y=1−xx, where x=α,β
y=1x−1⇒y+1=1x⇒x=1y+1
The equation whose 1−αα and 1−ββ is,
f(1x+1)=0⇒a+b(x+1)+c(x+1)2=0⇒cx2+(2c+b)x+(a+b+c)=0
Comparing with
px2+qx+r=0
cp=(2c+b)q=(a+b+c)r=k
Now,
cp=(2c+b)q
⇒qp=(2c+b)c⇒qp=2+bc∴k=2