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

1) $\frac{2}{3}$
2) $-\frac{2}{3}$
3) $\frac{1}{3}$
4) $-\frac{1}{3}$
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(1)

Explanation

$\beta=2 \alpha$

$3 \alpha=\frac{3 a-1}{a^{2}-5 a+3}$

$2 \alpha^{2}=\frac{2}{a^{2}-5 a+3}$

$\frac{(3 a-1)^{2}}{a\left(a^{2}-5 a+3\right)^{2}}=\frac{1}{a^{2}+5 a+3} \Rightarrow a=\frac{2}{3}$

by Diamond (74,124 points)

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