JEE Main 2016 (Online) 9th AprilQuestion 1
If the equations $x^{2}+bx-1=0$ and $x^{2}+x+b=0$ have a common root different from $-1$, then $|b|$ is equal to :
Quadratic Equation · Common Roots
Solution
Concept. Subtract the equations to express the common root in terms of $b$, then substitute back. Formulas used. (1) A common root $\alpha$ satisfies both equationsStep 1 - Subtract. $\alpha^2+b\alpha-1=0$ and $\alpha^2+\alpha+b=0$ give $(b-1)\alpha=b+1$, so $\alpha=\dfrac{b+1}{b-1}$. Step 2 - Substitute. Into $\alpha^2+\alpha+b=0$, multiplied by $(b-1)^2$: $$(b+1)^2+(b+1)(b-1)+b(b-1)^2=0\ \Longrightarrow\ b^3+3b=0.$$ Step 3 - Choose. $b=0$ gives $\alpha=-1$, which is excluded. So $b^2=-3$ and $|b|=\sqrt3$. Answer: (C)