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: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

Let the roots be α and 2 α .Sum of roots: α + 2 α = 3 α = - \frac{3 a - 1}{a ^{2} - 5 a + 3} ⟹ α = - \frac{3 a - 1}{3 ( a ^{2} - 5 a + 3 )} Product of roots: α (2 α) = 2 α^{2} = \frac{2}{a ^{2} - 5 a + 3} ⟹ α^{2} = \frac{1}{a ^{2} - 5 a + 3} Substituting α : (- \frac{3 a - 1}{3 ( a ^{2} - 5 a + 3 )})^{2} = \frac{1}{a ^{2} - 5 a + 3} \frac{( 3 a - 1 ) ^{2}}{9 ( a ^{2} - 5 a + 3 ) ^{2}} = \frac{1}{a ^{2} - 5 a + 3} (3 a - 1)^{2} = 9 (a^{2} - 5 a + 3) 9 a^{2} - 6 a + 1 = 9 a^{2} - 45 a + 27 39 a = 26 ⟹ a = \frac{26}{39} = \frac{2}{3}

Try it yourself before checking the explanation above.

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

\frac{2}{3}

- \frac{2}{3}

\frac{1}{3}

- \frac{1}{3}

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