[JEE Main 2003] The value of $a$ for which one root of $(a^2-5a+3)x^2+(3a-1)x+2=0$ is twice the other is
The value of $a$ for which one root of $(a^2-5a+3)x^2+(3a-1)x+2=0$ is twice the other is
(a) $-\frac13$
(b) $\frac23$
(c) $-\frac23$
(d) $\frac13$
1 Answer
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✓ Vidaara Team
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· 1mo ago
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Correct answer: (b) $\frac23$
Roots $r,2r$: $9(a^2-5a+3)=(3a-1)^2\Rightarrow-45a+27=-6a+1\Rightarrow a=\frac23$.
JEE Main 2003 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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