JEE Main 2021 (Online) 26th August Evening ShiftQuestion 1
Let $\lambda\neq0$ be in $\mathbf{R}$. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-x+2\lambda=0$, and $\alpha$ and $\gamma$ are the roots of the equation $3x^{2}-10x+27\lambda=0$, then $\frac{\beta\gamma}{\lambda}$ is equal to $\_\_\_\_\_\_$ .
Quadratic Equation · Common Roots
Answer$18$
Solution
Concept. Eliminating $\alpha^2$ gives $\alpha$ in terms of $\lambda$; then the other roots follow from the sums. Formulas used. (1) For $px^2+qx+r=0$: sum of roots $=-\dfrac qp$Step 1 - Eliminate. $(3\alpha^2-10\alpha+27\lambda)-3(\alpha^2-\alpha+2\lambda)=-7\alpha+21\lambda=0$, so $\alpha=3\lambda$. Step 2 - Find $\lambda$. $9\lambda^2-3\lambda+2\lambda=0$, so $\lambda=\frac19$ ($\lambda\ne0$) and $\alpha=\frac13$. Step 3 - Other roots. By (1): $\beta=1-\frac13=\frac23$ and $\gamma=\frac{10}{3}-\frac13=3$. Step 4 - Value. $\dfrac{\beta\gamma}{\lambda}=\dfrac{2}{1/9}=18$. Answer: 18