If x^2 - 11x + a and x^2 - 14x + 2a have a co | vedclass

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Question

(A) Let the common factor be $(x - \alpha)$.Then $\alpha^2 - 11\alpha + a = 0$  $(1)$And $\alpha^2 - 14\alpha + 2a = 0$  $(2)$Subtracting equation $(2

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) Let the common factor be $(x - \alpha)$.Then $\alpha^2 - 11\alpha + a = 0$  $(1)$And $\alpha^2 - 14\alpha + 2a = 0$  $(2)$Subtracting equation $(2)$ from equation $(1)$:$(\alpha^2 - 11\alpha + a) - (\alpha^2 - 14\alpha + 2a) = 0$$3\alpha - a = 0 \implies \alpha = \frac{a}{3}$Substitute $\alpha = \frac{a}{3}$ into equation $(1)$:$(\frac{a}{3})^2 - 11(\frac{a}{3}) + a = 0$$\frac{a^2}{9} - \frac{11a}{3} + a = 0$Multiply by $9$:$a^2 - 33a + 9a = 0$$a^2 - 24a = 0$$a(a - 24) = 0$Since $a$ must be non-zero for a non-trivial common factor,$a = 24$.

If $x^2 - 11x + a$ and $x^2 - 14x + 2a$ have a common factor,then $a = ......$

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