Multiple choice

If both the roots of the equations $k(6x^2 + 3) + rx + 2x^2 - 1 = 0$ and $6k(2x^2 + 1) + px + 4x^2 - 2 = 0$ are common, then $2r - p$ is equal to-

  1. $1$
  2. $-1$
  3. $2$
  4. $0$
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D Correct answer
Explanation

For the roots to be common, the coefficients must be proportional. Comparing the two equations, we find the ratio of coefficients leads to 2r - p = 0.

AI explanation

The equations can be rewritten in standard form as (6k + 2)x^2 + rx + (3k - 1) = 0 and (12k + 4)x^2 + px + (6k - 2) = 0. Notice that the coefficients of the second equation are exactly twice the coefficients of the first equation. Because two quadratic equations share both roots if and only if their corresponding coefficients are proportional, we must have p = 2r. Therefore, 2r - p equals 0.

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