Prove the following


Let $\lambda \neq 0$ be in $\mathbf{R}$. If $\alpha$ and $\beta$ are roots of the equation, $x^{2}-x+2 \lambda=0$ and $\alpha$ and $\gamma$ are the roots of the equation, $3 x^{2}-10 x+27 \lambda=0$, then $\frac{\beta \gamma}{\lambda}$ is equal to :

  1. (1) 27

  2. (2) 18

  3. (3) 9

  4. (4) 36

Correct Option: , 2


Since $\alpha$ is common root of $x^{2}-x+2 \lambda=0$ and

$3 x^{2}-10 x+27 \lambda=0$

$\therefore 3 \alpha^{2}-10 \alpha+27 \lambda=0$...(i)

$3 \alpha^{2}-3 \alpha+6 \lambda=0$..(ii)

$\therefore$ On subtract, we get $\alpha=3 \lambda$

Now, $\alpha \beta=2 \lambda \Rightarrow 3 \lambda \cdot \beta=2 \lambda \Rightarrow \beta=\frac{2}{3}$

$\Rightarrow \alpha+\beta=1 \Rightarrow 3 \lambda+\frac{2}{3}=1 \Rightarrow \lambda=\frac{1}{9}$ and

$\alpha \gamma=9 \lambda \Rightarrow 3 \lambda \cdot \gamma=9 \lambda \Rightarrow \gamma=3$

$\therefore \frac{\beta \gamma}{\lambda}=18$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now