Multiple choice

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

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

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

AI explanation

Let the roots be r and 2r. The sum of the roots gives 3r = -(3p - 1) / (p^2 - 5p + 3). The product of the roots gives 2r^2 = 2 / (p^2 - 5p + 3). Squaring the sum equation gives 9r^2 = (3p - 1)^2 / (p^2 - 5p + 3)^2, so 18 / (p^2 - 5p + 3) = (3p - 1)^2 / (p^2 - 5p + 3)^2. Simplifying yields 2(p^2 - 5p + 3) = 9p^2 - 6p + 1. Solving 7p^2 + 4p - 5 = 0 gives p = 2/3 or p = -5/7.

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