The value of a for which one root of the quad | vedclass

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(A) Let the roots of the quadratic equation be $\alpha$ and $2\alpha$.From the properties of roots,the sum of roots is $\alpha + 2\alpha = 3\alpha = -

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$\frac{2}{3}$

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(A) Let the roots of the quadratic equation be $\alpha$ and $2\alpha$.From the properties of roots,the sum of roots is $\alpha + 2\alpha = 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} = \frac{1 - 3a}{a^2 - 5a + 3}$.Thus,$\alpha = \frac{1 - 3a}{3(a^2 - 5a + 3)}$.The product of roots is $\alpha \cdot 2\alpha = 2\alpha^2 = \frac{2}{a^2 - 5a + 3}$.Substituting the value of $\alpha$ into the product equation:$2 \left[ \frac{1 - 3a}{3(a^2 - 5a + 3)} \right]^2 = \frac{2}{a^2 - 5a + 3}$.$2 \cdot \frac{(1 - 3a)^2}{9(a^2 - 5a + 3)^2} = \frac{2}{a^2 - 5a + 3}$.Canceling $2$ and one factor of $(a^2 - 5a + 3)$ from both sides:$\frac{(1 - 3a)^2}{9(a^2 - 5a + 3)} = 1$.$(1 - 3a)^2 = 9(a^2 - 5a + 3)$.$1 - 6a + 9a^2 = 9a^2 - 45a + 27$.$-6a + 45a = 27 - 1$.$39a = 26$.$a = \frac{26}{39} = \frac{2}{3}$.

The value of $a$ for which one root of the quadratic equation $(a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0$ is twice as large as the other,is

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