The quadratic equations x^2 - 6x + a = 0 and | vedclass

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(D) Let the roots of the equation $x^2 - 6x + a = 0$ be $\alpha$ and $4\beta$,and the roots of the equation $x^2 - cx + 6 = 0$ be $\alpha$ and $3\beta

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$2$

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(D) Let the roots of the equation $x^2 - 6x + a = 0$ be $\alpha$ and $4\beta$,and the roots of the equation $x^2 - cx + 6 = 0$ be $\alpha$ and $3\beta$,where $\alpha$ is the common root.From the sum and product of roots:For the first equation: $\alpha + 4\beta = 6$ and $4\alpha\beta = a$.For the second equation: $\alpha + 3\beta = c$ and $3\alpha\beta = 6$.From $3\alpha\beta = 6$,we get $\alpha\beta = 2$,so $\beta = \frac{2}{\alpha}$.Substitute $\beta = \frac{2}{\alpha}$ into the first equation: $\alpha + 4(\frac{2}{\alpha}) = 6$.$\alpha + \frac{8}{\alpha} = 6 \Rightarrow \alpha^2 - 6\alpha + 8 = 0$.Solving the quadratic equation: $(\alpha - 2)(\alpha - 4) = 0$.So,$\alpha = 2$ or $\alpha = 4$.If $\alpha = 2$,then $\beta = 1$. The roots of the first equation are $2$ and $4(1) = 4$. The roots of the second equation are $2$ and $3(1) = 3$. This satisfies the condition.If $\alpha = 4$,then $\beta = 0.5$,which is not an integer. Since the problem states the other roots are integers,$\alpha = 2$ is the correct common root.

The quadratic equations $x^2 - 6x + a = 0$ and $x^2 - cx + 6 = 0$ have one root in common. The other roots of the first and second equations are integers in the ratio $4 : 3$. Then the common root is

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