Consider the equation `x^(2) + 2x - n = 0`m where `n in N` and `n in [5, 100]`. The total number of different values of n so that the given equation has integral roots is

Options

A. 8
B. 3
C. 6
D. 4

Explanation

To solve the equation \(x^2 + 2x - n = 0\) for integral roots, we will follow these steps:

### Step 1: Identify the roots of the quadratic equation
The roots of the quadratic equation \(ax^2 + bx + c = 0\) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \(x^2 + 2x - n = 0\), we have \(a = 1\), \(b = 2\), and \(c = -n\). Plugging these values into the formula gives:
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-n)}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{-2 \pm \sqrt{4 + 4n}}{2} \]
\[ x = \frac{-2 \pm 2\sqrt{n + 1}}{2} \]
\[ x = -1 \pm \sqrt{n + 1} \]

### Step 2: Determine conditions for integral roots
For the roots \(x = -1 + \sqrt{n + 1}\) and \(x = -1 - \sqrt{n + 1}\) to be integers, \(\sqrt{n + 1}\) must also be an integer. This means that \(n + 1\) must be a perfect square.

### Step 3: Set up the equation for perfect squares
Let \(k^2 = n + 1\) for some integer \(k\). Therefore, we can express \(n\) as:
\[ n = k^2 - 1 \]

### Step 4: Determine the range for \(n\)
Given that \(n\) is a natural number and \(n\) is in the range \([5, 100]\), we can set up the inequalities:
\[ 5 \leq k^2 - 1 \leq 100 \]
Adding 1 to all parts of the inequality gives:
\[ 6 \leq k^2 \leq 101 \]

### Step 5: Find the integer values of \(k\)
Now we need to find the integer values of \(k\) such that \(k^2\) falls within the range \([6, 101]\). The smallest integer \(k\) such that \(k^2 \geq 6\) is \(k = 3\) (since \(3^2 = 9\)), and the largest integer \(k\) such that \(k^2 \leq 101\) is \(k = 10\) (since \(10^2 = 100\)).

### Step 6: List the possible values of \(k\)
The integer values of \(k\) that satisfy this condition are:
\[ k = 3, 4, 5, 6, 7, 8, 9, 10 \]
This gives us a total of \(10 - 3 + 1 = 8\) possible values for \(k\).

### Step 7: Calculate corresponding values of \(n\)
Now we can calculate the corresponding values of \(n\):
- For \(k = 3\), \(n = 3^2 - 1 = 8\)
- For \(k = 4\), \(n = 4^2 - 1 = 15\)
- For \(k = 5\), \(n = 5^2 - 1 = 24\)
- For \(k = 6\), \(n = 6^2 - 1 = 35\)
- For \(k = 7\), \(n = 7^2 - 1 = 48\)
- For \(k = 8\), \(n = 8^2 - 1 = 63\)
- For \(k = 9\), \(n = 9^2 - 1 = 80\)
- For \(k = 10\), \(n = 10^2 - 1 = 99\)

### Conclusion
The total number of different values of \(n\) such that the equation has integral roots is **8**.