Let common root be x
x2+bx−1=0 ...(1)
x2+x+b=0 ...(2)
(1)−(2), we get x=b−1b+1
Put x in Equation (1), we get
(b−1b+1)2+(b−1b+1)+b=0
⇒ (b+1)2+(b+1)(b−1)+b(b−1)2=0
⇒ b2+1+2b+b2−1+b(b2−2b+1)=0
⇒ 2b2+2b+b3−2b2+b=0
⇒ b3+3b=0
⇒ b(b2+3)=0⇒b=0 or b2=−3
When b=0 then common root x=−1
But the given common root x=−1
∴b2=−3
b=±3i
∣b∣=3