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

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(D) Let the common root be $\alpha$. Let the other roots be $\beta$ and $\gamma$ for the first and second equations respectively.Given $\beta$ and $\g

Vedclass · Accepted Answer

$2$

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(D) Let the common root be $\alpha$. Let the other roots be $\beta$ and $\gamma$ for the first and second equations respectively.Given $\beta$ and $\gamma$ are integers and $\frac{\beta}{\gamma} = \frac{4}{3}$.From the properties of roots:For $x^2 - 6x + a = 0$,$\alpha + \beta = 6$ and $\alpha\beta = a$.For $x^2 - cx + 6 = 0$,$\alpha + \gamma = c$ and $\alpha\gamma = 6$.Since $\beta = \frac{4}{3}\gamma$,substitute into the sum of roots for the first equation: $\alpha + \frac{4}{3}\gamma = 6 \implies 3\alpha + 4\gamma = 18$.From the product of roots for the second equation,$\gamma = \frac{6}{\alpha}$.Substitute $\gamma$ into the equation: $3\alpha + 4(\frac{6}{\alpha}) = 18 \implies 3\alpha^2 - 18\alpha + 24 = 0 \implies \alpha^2 - 6\alpha + 8 = 0$.Solving for $\alpha$: $(\alpha - 2)(\alpha - 4) = 0$,so $\alpha = 2$ or $\alpha = 4$.If $\alpha = 4$,then $\gamma = \frac{6}{4} = 1.5$ (not an integer).If $\alpha = 2$,then $\gamma = \frac{6}{2} = 3$ (an integer) and $\beta = \frac{4}{3} 	imes 3 = 4$ (an integer).Thus,the common root is $\alpha = 2$.

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

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