Question

The value of a such that  \[x^2 - 11x + a = 0 \text { and } x^2 - 14x + 2a = 0\] may have a common root is

Options

• 0

• 12

• 24

• 32

Answer 1

(a) and (c)

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

Therefore,

\[\alpha^2 - 11\alpha + a = 0\]     ... (1)

\[\alpha^2 - 14\alpha + 2a = 0\]  ... (2)

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

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

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

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

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

\[ \Rightarrow a^2 = 24a\]

\[ \Rightarrow a^2 - 24a = 0\]

\[ \Rightarrow a\left( a - 24 \right) = 0\]

\[ \Rightarrow a = 0 \text { or } a = 24\]

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.