Question

If the roots of the equation \( 32x^3 - 48x^2 + 22x - 3 = 0 \) are in arithmetic progression, then the square of the common difference of the roots is

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{16} \)
• C. \( \frac{1}{9} \)
• D. \( \frac{1}{25} \)

Hint:

When roots of a cubic are in A.P., the middle term is simply \( \frac{\text{Sum of roots}}{3} \). Finding this root first simplifies the problem significantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

For a cubic equation \( Ax^3 + Bx^2 + Cx + D = 0 \) with roots in Arithmetic Progression (A.P.), we can assume the roots to be \( \alpha-d, \alpha, \alpha+d \).
Step 2: Key Formula or Approach:

Sum of roots \( = -B/A \). Product of roots \( = -D/A \).
Step 3: Detailed Explanation:

Let the roots be \( \alpha-d, \alpha, \alpha+d \). Sum of roots: \[ (\alpha-d) + \alpha + (\alpha+d) = -\frac{-48}{32} \] \[ 3\alpha = \frac{48}{32} = \frac{3}{2} \] \[ \alpha = \frac{1}{2} \] So, the middle root is \( \frac{1}{2} \). Since it is a root, it satisfies the equation (verified: \( 32(1/8) - 48(1/4) + 22(1/2) - 3 = 4 - 12 + 11 - 3 = 0 \)). Product of roots: \[ (\alpha-d)(\alpha)(\alpha+d) = -\frac{-3}{32} \] \[ \alpha(\alpha^2 - d^2) = \frac{3}{32} \] Substitute \( \alpha = \frac{1}{2} \): \[ \frac{1}{2} \left( \left(\frac{1}{2}\right)^2 - d^2 \right) = \frac{3}{32} \] Multiply by 2: \[ \frac{1}{4} - d^2 = \frac{6}{32} = \frac{3}{16} \] Solve for \( d^2 \): \[ d^2 = \frac{1}{4} - \frac{3}{16} \] \[ d^2 = \frac{4}{16} - \frac{3}{16} = \frac{1}{16} \]
Step 4: Final Answer:

The square of the common difference is \( \frac{1}{16} \).