Question

If two roots of the equation \( x^5 - 9x^4 + 27x^3 - 23x^2 - 24x + 36 = 0 \) are repeated roots of multiplicity 2 and all the roots are integers then the sum of the squares of all different roots of the equation is:

• A. \( 14 \)
• B. \( 29 \)
• C. \( 26 \)
• D. \( 21 \)

Hint:

When multiplicity is given, the product of roots often simplifies to a form like \( k^2 \cdot c \). Start by testing small perfect square factors of the constant term to find the repeated roots.

Answer 1

The Correct Option is A

Concept: For a polynomial equation with integer roots, we use Vieta's formulas and the factor theorem.
• Let the roots be \( a, a, b, b, c \).
• Sum of roots: \( 2a + 2b + c = 9 \).
• Product of roots: \( a^2 \cdot b^2 \cdot c = -36 \).

Step 1: Identifying the roots using integer properties.
Since \( a^2b^2c = -36 \), \( a^2b^2 \) must be a perfect square divisor of 36. Possible values for \( \{|a|, |b|\} \) are \( \{1, 2\}, \{1, 3\}, \{2, 3\} \). If \( a = 2 \) and \( b = 3 \), then \( a^2b^2 = 4 \times 9 = 36 \). This forces \( c = -1 \).

Step 2: Checking the sum of roots.
Check if \( 2(2) + 2(3) + (-1) = 9 \): \[ 4 + 6 - 1 = 9 \] This matches the coefficient of \( x^4 \). Thus, the roots are \( 2, 2, 3, 3, -1 \).

Step 3: Calculating the sum of squares of different roots.
The distinct (different) roots are \( \{2, 3, -1\} \). \[ \text{Sum of squares} = 2^2 + 3^2 + (-1)^2 = 4 + 9 + 1 = 14 \]