Multiple choice

If the equations K(6x2 + 3) + rx + 2x2 - 1 = 0 and 6K(2x2 + 1) + Px + 4x2 -2 = 0 have both the roots common, then (2r - P) is

  1. K

  2. 1

  3. 0

  4. 3K
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For two quadratic equations to have both roots common, the ratios of their coefficients must be equal. Dividing the second equation by 2 gives 3K(x^2 + 0.5) + (P/2)x + 2x^2 - 1 = 0, which simplifies to (3K+2)x^2 + (P/2)x + (1.5K-1) = 0. Comparing this to the first equation K(6x^2+3) + rx - 1 = 0, or (6K+2)x^2 + rx + (3K-1) = 0, we find the ratios match when 2r = P, making 2r - P = 0.

AI explanation

Rearrange the first equation to (6K + 2)x^2 + rx + (3K - 1) = 0 and the second to (12K + 4)x^2 + Px + (6K - 2) = 0. Notice that the coefficients of the second equation are exactly twice those of the first, since 12K + 4 = 2(6K + 2) and 6K - 2 = 2(3K - 1). Therefore, the ratio of the x coefficients must also be equal, giving P = 2r, which means 2r - P = 0.

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