Complex Numbers and Quadratic Equations — MATHS STD 11 Science — Question

2. -1

Solution:

Let a be the common roots of the equations $\text{x}^2+2\text{x}+3\lambda=0$ and $2\text{x}^2+3\text{x}+5\lambda=0$

Therefore

$\alpha^2+2\text{a}+3\lambda=0\ ...(1)$

$2\alpha^2+3\alpha+5\lambda=0\ ...(2)$

Solving (1) and (2) by cross multiplication, we get

$\frac{\alpha^2}{10\lambda-9\lambda}=\frac{\alpha}{6\lambda-5\lambda}=\frac{1}{3-4}$

$\Rightarrow\text{a}^2=-\lambda,\alpha=-\lambda$

$\Rightarrow-\lambda=\lambda^2$

$\Rightarrow\lambda=-1$