If the equations x^2 - 11x + a = 0 and x^2 - | vedclass

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(C) Let $(x - \alpha)$ be the common factor. Then $x = \alpha$ is a root of both equations. $\alpha^2 - 11\alpha + a = 0$  --- $(1)$ $\alpha^2 - 14\al

Vedclass · Accepted Answer

$(x - 8)$

Vedclass · Answer

(C) Let $(x - \alpha)$ be the common factor. Then $x = \alpha$ is a root of both equations. $\alpha^2 - 11\alpha + a = 0$  --- $(1)$ $\alpha^2 - 14\alpha + 2a = 0$ --- $(2)$ Subtracting $(1)$ from $(2)$: $(\alpha^2 - 14\alpha + 2a) - (\alpha^2 - 11\alpha + a) = 0$ $-3\alpha + a = 0 \implies \alpha = \frac{a}{3}$ Substitute $\alpha = \frac{a}{3}$ into equation $(1)$: $(\frac{a}{3})^2 - 11(\frac{a}{3}) + a = 0$ $\frac{a^2}{9} - \frac{11a}{3} + a = 0$ $\frac{a^2 - 33a + 9a}{9} = 0 \implies a^2 - 24a = 0$ $a(a - 24) = 0$ Since $a 
eq 0$,we have $a = 24$. Substituting $a = 24$ into the equations: $x^2 - 11x + 24 = (x - 8)(x - 3)$ $x^2 - 14x + 48 = (x - 8)(x - 6)$ The common factor is $(x - 8)$.

If the equations $x^2 - 11x + a = 0$ and $x^2 - 14x + 2a = 0$ have a common factor and $a \neq 0$,then what is the common factor?

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