Multiple choice

If one root of the equation $ 8x^{2}-6x+k=0 $ is the square of the other, then the value of $k$ is

  1. $0, 3$
  2. $-1, 27$
  3. $0, -2$
  4. $1, -27$
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D Correct answer
Explanation

Let the roots be a and a^2. From the sum of roots, a + a^2 = 6/8 = 3/4. From the product, a * a^2 = a^3 = k/8. Solving a + a^2 = 3/4 gives a = 1/2 or a = -3/2. If a = 1/2, a^3 = 1/8, so k/8 = 1/8, k = 1. If a = -3/2, a^3 = -27/8, so k/8 = -27/8, k = -27.

AI explanation

Let the roots be a and a^2. The product of roots is k/8, so a^3 = k/8. The sum of roots is 6/8, giving a + a^2 = 3/4. Multiplying by 4 gives 4a^2 + 4a - 3 = 0, which factors as (2a-1)(2a+3)=0. If a = 1/2, then k = 8(1/8) = 1. If a = -3/2, then k = 8(-27/8) = -27. The values of k are 1, -27.

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