Question

The number of all common roots of the equation \( x^4 - 10x^3 + 37x^2 - 60x + 36 = 0 \) and the transformed equation of it obtained by increasing any two distinct roots of it by 1, keeping the other two roots fixed, is

• A. 1
• B. 3
• C. 4
• D. 2

Hint:

When dealing with "common roots" of higher-degree polynomials with repeated roots, clarify if the question implies distinct values or total count (multiplicity). In competitive exams, if options include the multiplicity sum (here 3), it is the intended answer over distinct count (here 2).

Answer 1

The Correct Option is B

Step 1: Find roots of the original equation:

Let \( P(x) = x^4 - 10x^3 + 37x^2 - 60x + 36 = 0 \). Check for integer roots (factors of 36). \( P(2) = 16 - 80 + 148 - 120 + 36 = 200 - 200 = 0 \). So \( 2 \) is a root. \( P(3) = 81 - 270 + 333 - 180 + 36 = 450 - 450 = 0 \). So \( 3 \) is a root. Checking multiplicity: Dividing \( P(x) \) by \( (x-2)(x-3) = x^2 - 5x + 6 \). We find that \( P(x) = (x^2 - 5x + 6)^2 = (x-2)^2 (x-3)^2 \). The roots of the original equation are \( S_1 = \{2, 2, 3, 3\} \). The distinct roots are 2 and 3.
Step 2: Form the transformed equation:

Rule: "Increasing any two distinct roots of it by 1, keeping the other two roots fixed". The distinct roots available to increase are 2 and 3. We take one instance of root '2' and increase it by 1 \( \to 3 \). We take one instance of root '3' and increase it by 1 \( \to 4 \). The other two roots (one '2' and one '3') remain fixed. The new set of roots is: 1. Fixed '2' \( \to 2 \) 2. Fixed '3' \( \to 3 \) 3. Increased '2' \( \to 3 \) 4. Increased '3' \( \to 4 \) Roots of transformed equation: \( S_2 = \{2, 3, 3, 4\} \).
Step 3: Find common roots:

We compare the multisets of roots: Original: \{2, 2, 3, 3\} Transformed: \{2, 3, 3, 4\} Common roots (values that satisfy both):
- \( x = 2 \) is a root in both (appears twice in original, once in transformed). It is a common root.
- \( x = 3 \) is a root in both (appears twice in original, twice in transformed). It is a common root. The question asks for "The number of all common roots". This typically sums the multiplicities of the intersection. Intersection of multisets: \{2, 3, 3\}. - One '2' - Two '3's Total count = \( 1 + 2 = 3 \).
Step 4: Final Answer:

The number of common roots is 3.