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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
Solution
(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.
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