If the expression $x^2-11 x+a$ and $x^2-14 x+2 a$ must have a common factor and $\mathrm{a} \neq 0$, then, the common factor is

Question

If the expression $x^2-11 x+a$ and $x^2-14 x+2 a$ must have a common factor and $\mathrm{a} \neq 0$, then, the common factor is, $(x-3)$, $(x-6)$, $(x-8)$, None of these

MARKS App · Accepted Answer

Here Let $x-\alpha$ is the common factor then $\mathrm{x}=\alpha$ is root of the corresponding equation $\therefore \alpha^2-11 \alpha+a=0$ $\alpha^2-14 \alpha+2 \mathrm{a}=0$ Subtracting $3 \alpha-a=0 \Rightarrow \alpha=a / 3$ Hence $\frac{a^2}{9}-11 \frac{a}{3}+a=0, a=0$ or $a=24$ since $\mathrm{a} \neq 0, \mathrm{a}=24$ $\therefore$ the common factor of $\left\{\begin{array}{l}x^2-11 x+24 \ x^2-14 x+48\end{array}\right.$ is clearly $x -8$

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