Question:medium

The value of \(a \ge b\) for which the sum of the cubes of the roots of \(x^2 - (a - 2)x + (a - 3) = 0\) assumes the least value, is

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Sum of cubes = \((\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\).
Updated On: Jun 16, 2026
  • 3
  • 4
  • 5
  • None of the above
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The Correct Option is A

Solution and Explanation

To find the value of \(a \ge b\) such that the sum of the cubes of the roots of the quadratic equation \(x^2 - (a - 2)x + (a - 3) = 0\) is minimized, we start by understanding the problem using Vieta's formulas. Let's denote the roots of the equation by \(\alpha\) and \(\beta\).

The quadratic equation is given as:

\[ x^2 - (a - 2)x + (a - 3) = 0 \]

By Vieta's formulas, we know:

  • The sum of the roots: \(\alpha + \beta = a - 2\)
  • The product of the roots: \(\alpha \beta = a - 3\)

We need to find the sum of the cubes of the roots, which is \(\alpha^3 + \beta^3\). The formula for \(\alpha^3 + \beta^3\) in terms of \(\alpha + \beta\) and \(\alpha \beta\) is:

\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]

Using the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\), we find:

\[ \alpha^2 + \beta^2 = (a - 2)^2 - 2(a - 3) \]

\[ \alpha^2 + \beta^2 = a^2 - 4a + 4 - 2a + 6 = a^2 - 6a + 10 \]

Thus, \(\alpha^3 + \beta^3\) becomes:

\[ \alpha^3 + \beta^3 = (a - 2)((a^2 - 6a + 10) - (a - 3)) \]

\[ = (a - 2)(a^2 - 6a + 10 - a + 3) \]

\[ = (a - 2)(a^2 - 7a + 13) \]

To find the minimum value of \((a - 2)(a^2 - 7a + 13)\), we can examine the critical points by differentiating it with respect to \(a\) and setting it equal to zero if necessary.

However, we can bypass this by examining integer options provided:

  1. a = 3: \((3 - 2)(3^2 - 7\times3 + 13) = 1 \times (9 - 21 + 13) = 1\)
  2. a = 4: \((4 - 2)(4^2 - 7\times4 + 13) = 2 \times (16 - 28 + 13) = 2\)
  3. a = 5: \((5 - 2)(5^2 - 7\times5 + 13) = 3 \times (25 - 35 + 13) = 9\)

The minimum is clearly at \(a = 3\), where the evaluated expression equals 1. Thus, the correct answer is 3.

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