The correct option is C b=−c
Let the roots of the equation f(x)=ax2+bx+c=0 be α,β
Now, the equation with roots α−1,β−1 is,
f(x+1)=0⇒a(x+1)2+b(x+1)+c=0
⇒ax2+(2a+b)x+(a+b+c)=0
Comparing to the equation
2x2+8x+2=0
So,
a2=(2a+b)8=a+b+c2=k
Where is some constant,
⇒a=2k⇒2a+b=8k⇒b=4k⇒a+b+c=2k⇒c=−4k
Therefore,
b=−c, b=2a, c=−2a
a:b:c=1:2:−2