For how many values a in mathbb{C},do the equ | vedclass

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(C) Let $\alpha$ be the common root of the given equations:$\alpha^2 - 8\alpha + 7 = 0$  $(i)$$\alpha^2 - 2a\alpha + 49 = 0$  $(ii)$Subtracting $(ii)$

Vedclass · Accepted Answer

$2$

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(C) Let $\alpha$ be the common root of the given equations:$\alpha^2 - 8\alpha + 7 = 0$  $(i)$$\alpha^2 - 2a\alpha + 49 = 0$  $(ii)$Subtracting $(ii)$ from $(i)$:$(\alpha^2 - 8\alpha + 7) - (\alpha^2 - 2a\alpha + 49) = 0$$(2a - 8)\alpha - 42 = 0$$2(a - 4)\alpha = 42$$\alpha = \frac{21}{a - 4}$Substituting $\alpha$ into $(i)$:$(\frac{21}{a - 4})^2 - 8(\frac{21}{a - 4}) + 7 = 0$Multiplying by $(a - 4)^2$:$441 - 168(a - 4) + 7(a - 4)^2 = 0$$441 - 168a + 672 + 7(a^2 - 8a + 16) = 0$$7a^2 - 168a - 56a + 1113 + 112 = 0$$7a^2 - 224a + 1225 = 0$Dividing by $7$:$a^2 - 32a + 175 = 0$$(a - 7)(a - 25) = 0$Thus,$a = 7$ or $a = 25$.There are $2$ possible values for $a$.

For how many values $a \in \mathbb{C}$,do the equations $x^2-8x+7=0$ and $x^2-2ax+49=0$ have a common root?

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