A = {x : x satisfies `x^3-11x^2 + 39x-45 = 0`} B={x:5x-6,3x+1,x-1 are lengths of sides of `Delta`}, then `n(AnnB)` is

Answer

To solve the problem, we need to find the intersection of two sets A and B, and then determine the number of elements in that intersection.

### Step 1: Define Set A
Set A is defined as:
\[ A = \{ x : x \text{ satisfies } x^3 - 11x^2 + 39x - 45 = 0 \} \]

We will first find the roots of the polynomial equation.

### Step 2: Find Roots of the Polynomial
To find the roots of the polynomial \( x^3 - 11x^2 + 39x - 45 = 0 \), we can use the Rational Root Theorem or synthetic division.

Let's test \( x = 3 \):
\[ 3^3 - 11(3^2) + 39(3) - 45 = 27 - 99 + 117 - 45 = 0 \]
So, \( x = 3 \) is a root.

### Step 3: Factor the Polynomial
Now we can factor the polynomial using \( x - 3 \):
\[ x^3 - 11x^2 + 39x - 45 = (x - 3)(Ax^2 + Bx + C) \]
Using synthetic division or polynomial long division, we find:
\[ x^3 - 11x^2 + 39x - 45 = (x - 3)(x^2 - 8x + 15) \]
Next, we can factor \( x^2 - 8x + 15 \):
\[ x^2 - 8x + 15 = (x - 3)(x - 5) \]
Thus, the complete factorization is:
\[ x^3 - 11x^2 + 39x - 45 = (x - 3)^2(x - 5) \]
The roots are \( x = 3 \) (with multiplicity 2) and \( x = 5 \).

### Step 4: Define Set B
Set B is defined as:
\[ B = \{ x : 5x - 6, 3x + 1, x - 1 \text{ are lengths of sides of a triangle} \} \]

### Step 5: Apply Triangle Inequality
For three lengths to form a triangle, they must satisfy the triangle inequalities:
1. \( (5x - 6) + (3x + 1) > (x - 1) \)
2. \( (5x - 6) + (x - 1) > (3x + 1) \)
3. \( (3x + 1) + (x - 1) > (5x - 6) \)

#### Inequality 1:
\[ (5x - 6) + (3x + 1) > (x - 1) \implies 8x - 5 > x - 1 \implies 7x > 4 \implies x > \frac{4}{7} \]

#### Inequality 2:
\[ (5x - 6) + (x - 1) > (3x + 1) \implies 6x - 7 > 3x + 1 \implies 3x > 8 \implies x > \frac{8}{3} \]

#### Inequality 3:
\[ (3x + 1) + (x - 1) > (5x - 6) \implies 4x > 5x - 7 \implies -x > -7 \implies x < 7 \]

### Step 6: Combine Inequalities for Set B
From the inequalities, we have:
1. \( x > \frac{8}{3} \)
2. \( x < 7 \)

Thus, set B can be defined as:
\[ B = \{ x : \frac{8}{3} < x < 7 \} \]

### Step 7: Find Intersection of Sets A and B
Set A has elements \( \{3, 5\} \) and set B has elements in the interval \( \left(\frac{8}{3}, 7\right) \).

- \( 3 \) is in the interval \( \left(\frac{8}{3}, 7\right) \).
- \( 5 \) is also in the interval \( \left(\frac{8}{3}, 7\right) \).

Thus, the intersection \( A \cap B = \{3, 5\} \).

### Step 8: Count the Number of Elements in the Intersection
The number of elements in the intersection \( n(A \cap B) = 2 \).

### Final Answer
The number of elements in the set \( A \cap B \) is:
\[ n(A \cap B) = 2 \]