The value of $a$ such that $x^{2}-11 x+a=0$ and $x^{2}-14 x+2 a=0$ may have a common root is
(a) 0
(b) 12
(c) 24
(d) 32
(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.
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