Question

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

• A. 2/3
• B. – 2/3
• C. 1/3
• D. – 1/3

Answer 1

Answer: (A)

2/3

Answer 2

Let the roots are α and 2α

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

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

⇒ $2\left\lbrack \frac{1}{9}\frac{(1 - 3a)^{2}}{(a^{2} - 5a + 3)^{2}} \right\rbrack = \frac{2}{a^{2} - 5a + 3}$ ⇒ $\frac{(1 - 3a)^{2}}{a^{2} - 5a + 3} = 9$

⇒ $9a^{2} - 45a + 27 = 1 + 9a^{2} - 6a$ ⇒ $39a = 26$⇒ $a = 2/3$