JEE PYQ: Quadratic Equation Question 20
Question 20 - 2020 (04 Sep Shift 2)
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, $3x^2 - 10x + 27\lambda = 0$, then $\frac{\beta\gamma}{\lambda}$ is equal to:
(1) 27
(2) 18
(3) 9
(4) 36
Show Answer
Answer: (2)
Solution
Since $\alpha$ is common root of $x^2 - x + 2\lambda = 0$ and $3x^2 - 10x + 27\lambda = 0$
$\therefore 3\alpha^2 - 10\alpha + 27\lambda = 0$ …(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
$\alpha\gamma = 9\lambda \Rightarrow 3\lambda \cdot \gamma = 9\lambda \Rightarrow \gamma = 3$
$\therefore \frac{\beta\gamma}{\lambda} = 18$
BETA
