Multiple choice

If the roots of the equation $4\ {x}^{3}-12\ {x}^{2}+\ {11}\ {x}+\ {k}=0$ are in A.P. Then $\ {k}=$

  1. $-3$
  2. $1$
  3. $2$
  4. $3$
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A Correct answer
Explanation

If roots of ax^3 + bx^2 + cx + d = 0 are in AP, the middle root is -b/(3a). Here a=4, b=-12, so middle root = -(-12)/(3*4) = 1. Since 1 is a root, 4(1)^3 - 12(1)^2 + 11(1) + k = 0. 4 - 12 + 11 + k = 0. 3 + k = 0, so k = -3.

AI explanation

Let the three roots of the cubic equation in arithmetic progression be a - d, a, and a + d. Since the coefficient of x^2 is -12, the sum of the roots 3a = -(-12)/4 = 3, meaning a = 1. The sum of the product of the roots taken two at a time equals 11/4, so we write (a-d)a + a(a+d) + (a-d)(a+d) = 11/4. Substituting a = 1 gives 1 - d + 1 + d + 1 - d^2 = 11/4, which simplifies to 3 - d^2 = 11/4, meaning d^2 = 1/4. The product of the roots equals -k/4, so (1-d)(1)(1+d) = -k/4, and substituting d^2 = 1/4 gives 1 - 1/4 = -k/4. Solving this yields 3/4 = -k/4, which means k = -3.

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