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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

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Solution

## (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, $\frac{{\mathrm{\alpha }}^{2}}{-22a+14a}=\frac{\mathrm{\alpha }}{a-2a}=\frac{1}{-14+11}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{\alpha }}^{2}=\frac{-22a+14a}{-14+11},\mathrm{\alpha }=\frac{a-2a}{-14+11}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{\alpha }}^{2}=\frac{-8a}{-3}=\frac{8a}{3},\mathrm{\alpha }=\frac{-a}{-3}=\frac{a}{3}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{\mathrm{a}}{3}\right)}^{2}=\frac{8a}{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=24a\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-24a=0\phantom{\rule{0ex}{0ex}}⇒a\left(a-24\right)=0\phantom{\rule{0ex}{0ex}}⇒a=0\mathrm{or}a=24$ 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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