For what value of a is one root of the quadra | vedclass

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(D) Let the roots be $\alpha$ and $2\alpha$.From the sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} \Rightarrow \alpha = -\f

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$2$

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(D) Let the roots be $\alpha$ and $2\alpha$.From the sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} \Rightarrow \alpha = -\frac{3a - 1}{3(a^2 - 5a + 3)}$.From the product of roots: $\alpha 	imes 2\alpha = 2\alpha^2 = \frac{2}{a^2 - 5a + 3} \Rightarrow \alpha^2 = \frac{1}{a^2 - 5a + 3}$.Squaring the sum of roots expression: $\alpha^2 = \frac{(3a - 1)^2}{9(a^2 - 5a + 3)^2}$.Equating the two expressions for $\alpha^2$: $\frac{1}{a^2 - 5a + 3} = \frac{(3a - 1)^2}{9(a^2 - 5a + 3)^2}$.Since $a^2 - 5a + 3 
eq 0$,we have $9(a^2 - 5a + 3) = (3a - 1)^2$.$9a^2 - 45a + 27 = 9a^2 - 6a + 1$.$-39a = -26 \Rightarrow a = \frac{26}{39} = \frac{2}{3}$.

For what value of $a$ is one root of the quadratic equation $(a^2 - 5a + 3) x^2 + (3a - 1) x + 2 = 0$ twice the other (in $/3$)?

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