The value of a such that x

Solution:

(a) and (c)

Let $\alpha$ be the common roots of the equations $x^{2}-11 x+a=0$ and $x^{2}-14 x+2 a=0$.

Therefore,

$\alpha^{2}-11 \alpha+a=0$        ...(1)

$\alpha^{2}-14 \alpha+2 a=0$    ...(2)

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

$\frac{\alpha^{2}}{-22 a+14 a}=\frac{\alpha}{a-2 a}=\frac{1}{-14+11}$

$\Rightarrow \alpha^{2}=\frac{-22 a+14 a}{-14+11}, \alpha=\frac{a-2 a}{-14+11}$

$\Rightarrow \alpha^{2}=\frac{-8 a}{-3}=\frac{8 a}{3}, \alpha=\frac{-a}{-3}=\frac{a}{3}$

$\Rightarrow\left(\frac{a}{3}\right)^{2}=\frac{8 a}{3}$

$\Rightarrow a^{2}=24 a$

$\Rightarrow a^{2}-24 a=0$

$\Rightarrow a(a-24)=0$

$\Rightarrow a=0$ or $a=24$

Disclaimer: The solution given in the book is incomplete. The solution is created according to the question given in the book and both the options are correct.