The number of real solutions of the equation `4x^(99) +5x^(98) + 4x^(97) + 5x^(96) + …..+ 4x + 5 =0` is (1) 1 (2) 5 (3) 7 (4) 97

Answer

To solve the equation \( 4x^{99} + 5x^{98} + 4x^{97} + 5x^{96} + \ldots + 4x + 5 = 0 \), we will analyze the structure of the polynomial and determine the number of real solutions.

### Step-by-Step Solution:

1. **Rewrite the Polynomial**:
The given polynomial can be grouped based on the coefficients of \(4\) and \(5\):
\[ 4(x^{99} + x^{97} + x^{95} + \ldots + x) + 5(x^{98} + x^{96} + x^{94} + \ldots + 1) = 0 \]

2. **Identify the Series**:
The terms \(x^{99} + x^{97} + x^{95} + \ldots + x\) form a geometric series with the first term \(x\) and the common ratio \(x^2\). The number of terms is \(50\) (since the highest power is \(99\) and we decrease by \(2\) each time).
The sum of this series can be expressed as:
\[ S_1 = x \frac{1 - x^{50}}{1 - x^2} \quad \text{(for } x \neq 1\text{)} \]

Similarly, the terms \(x^{98} + x^{96} + x^{94} + \ldots + 1\) also form a geometric series with the first term \(1\) and the common ratio \(x^2\). The number of terms is \(50\):
\[ S_2 = \frac{1 - x^{50}}{1 - x^2} \quad \text{(for } x \neq 1\text{)} \]

3. **Substituting Back**:
Substitute \(S_1\) and \(S_2\) back into the polynomial:
\[ 4 \cdot S_1 + 5 \cdot S_2 = 0 \]
\[ 4 \left( x \frac{1 - x^{50}}{1 - x^2} \right) + 5 \left( \frac{1 - x^{50}}{1 - x^2} \right) = 0 \]
Factor out the common term:
\[ \frac{(1 - x^{50})}{1 - x^2} (4x + 5) = 0 \]

4. **Setting Each Factor to Zero**:
This gives us two cases to solve:
- Case 1: \(1 - x^{50} = 0\) which implies \(x^{50} = 1\). The solutions are \(x = 1\) and \(x = -1\) (as \(x^{50} = 1\) has \(50\) solutions, but only \(x = 1\) and \(x = -1\) are real).
- Case 2: \(4x + 5 = 0\) which gives \(x = -\frac{5}{4}\).

5. **Counting Real Solutions**:
From Case 1, we have two real solutions \(x = 1\) and \(x = -1\). From Case 2, we have one additional real solution \(x = -\frac{5}{4}\). Thus, the total number of distinct real solutions is:
\[ 2 + 1 = 3 \]

### Final Answer:
The number of real solutions of the equation is \(3\).