Question

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

Answer 1

(a) and (c)

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

Therefore,

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

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

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

$$\begin{gathered} \frac{{\mathrm{\alpha }}^2}{-22a+14a}=\frac{\mathrm{\alpha }}{a-2a}=\frac{1}{-14+11} \\ \Rightarrow {\mathrm{\alpha }}^2=\frac{-22a+14a}{-14+11},\mathrm{\alpha }=\frac{a-2a}{-14+11} \\ \Rightarrow {\mathrm{\alpha }}^2=\frac{-8a}{-3}=\frac{8a}{3},\mathrm{\alpha }=\frac{-a}{-3}=\frac{a}{3} \\ \Rightarrow {\left(\frac{\mathrm{a}}{3}\right)}^2=\frac{8a}{3} \\ \Rightarrow a^2=24a \\ \Rightarrow a^2-24a=0 \\ \Rightarrow a\left(a-24\right)=0 \\ \Rightarrow a=0\mathrm{or}a=24 \end{gathered}$$

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.