If the equations x

(b) −1

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

Therefore,

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

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

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

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

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

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

$\Rightarrow \lambda=-1$