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

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(A) Let the roots be $\alpha$ and $2\alpha$.From the relation between roots and coefficients:Sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{3a - 1

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) Let the roots be $\alpha$ and $2\alpha$.From the relation between roots and coefficients:Sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} = \frac{1 - 3a}{a^2 - 5a + 3} \implies \alpha = \frac{1 - 3a}{3(a^2 - 5a + 3)}$.Product of roots: $\alpha \cdot 2\alpha = 2\alpha^2 = \frac{2}{a^2 - 5a + 3} \implies \alpha^2 = \frac{1}{a^2 - 5a + 3}$.Substituting $\alpha$ from the sum equation into the product equation:$\left[\frac{1 - 3a}{3(a^2 - 5a + 3)}\right]^2 = \frac{1}{a^2 - 5a + 3}$.$\frac{(1 - 3a)^2}{9(a^2 - 5a + 3)^2} = \frac{1}{a^2 - 5a + 3}$.$(1 - 3a)^2 = 9(a^2 - 5a + 3)$.$1 - 6a + 9a^2 = 9a^2 - 45a + 27$.$39a = 26$.$a = \frac{26}{39} = \frac{2}{3}$.

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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