Multiple choice

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

  1. $\displaystyle \frac{2}{3}$
  2. $\displaystyle \frac{1}{3}$
  3. $\displaystyle \frac{-2}{3}$
  4. $\displaystyle \frac{-1}{3}$
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A Correct answer
Explanation

Let roots be r and 2r. Sum = 3r = -(3a-1)/(a^2-5a+3). Product = 2r^2 = 2/(a^2-5a+3). From product, r^2 = 1/(a^2-5a+3). Squaring the sum: 9r^2 = (3a-1)^2 / (a^2-5a+3)^2. 9/(a^2-5a+3) = (3a-1)^2 / (a^2-5a+3)^2. 9 = (3a-1)^2 / (a^2-5a+3). 9(a^2-5a+3) = 9a^2 - 6a + 1. 9a^2 - 45a + 27 = 9a^2 - 6a + 1. 39a = 26. a = 26/39 = 2/3.

AI explanation

Let the roots be r and 2r, so their sum is 3r and product is 2r^2. Using the relationships between roots and coefficients for the equation (a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0, we set 3r = -(3a - 1) / (a^2 - 5a + 3) and 2r^2 = 2 / (a^2 - 5a + 3). From the second equation, r^2 = 1 / (a^2 - 5a + 3). Substituting r = 1 / (3r) from the first equation into r^2 gives 1 / (3r^2) = (1 - 3a) / (a^2 - 5a + 3). Replacing r^2 yields 1 / 3 = 1 - 3a. Solving this gives 3a = 2, so a equals 2/3.

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