Which one of the following is not true? The quadratic equation `x^(2) - 2x - a = 0, a ne 0`,

Options

A. cannot have a real root if `a lt -1`
B. may not have a rational root even if a is a perfect square
C. cannot have an integral root if `n^(2)-1 lt a lt n^(2) + 2n`, where n = 0, 1, 2,......
D. none of these

Explanation

To determine which statement about the quadratic equation \( x^2 - 2x - a = 0 \) (where \( a \neq 0 \)) is not true, we will analyze each option step by step.

### Step 1: Identify the Discriminant
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by the formula:
\[ D = b^2 - 4ac \]
For our equation \( x^2 - 2x - a = 0 \), we have:
- \( a = 1 \)
- \( b = -2 \)
- \( c = -a \)

Thus, the discriminant becomes:
\[ D = (-2)^2 - 4(1)(-a) = 4 + 4a = 4(1 + a) \]

### Step 2: Analyze Each Option

#### Option 1: Cannot have a real root if \( a < -1 \)
For the quadratic to have real roots, the discriminant must be non-negative:
\[ D \geq 0 \implies 4(1 + a) \geq 0 \implies 1 + a \geq 0 \implies a \geq -1 \]
If \( a < -1 \), then \( D < 0 \), meaning the equation cannot have real roots. **This option is true.**

#### Option 2: May not have a rational root even if \( a \) is a perfect square
Let \( a = k^2 \) (where \( k \) is an integer). Then:
\[ D = 4(1 + k^2) \]
Since \( 1 + k^2 \) is always positive, \( D \) is positive, indicating that there are real roots. However, whether the roots are rational depends on whether \( D \) is a perfect square:
\[ D = 4(1 + k^2) \text{ is not necessarily a perfect square.} \]
Thus, it may not have rational roots even if \( a \) is a perfect square. **This option is true.**

#### Option 3: Cannot have an integral root if \( n^2 - 1 < a < n^2 + 2n \) for \( n = 0, 1, 2, \ldots \)
To check this, we set \( x = \alpha \) (an integer) in the equation:
\[ \alpha^2 - 2\alpha - a = 0 \implies a = \alpha^2 - 2\alpha \]
The expression \( \alpha^2 - 2\alpha \) can be analyzed:
- If \( n^2 - 1 < a < n^2 + 2n \), we can check if there exists an integer \( \alpha \) such that \( a = \alpha^2 - 2\alpha \) falls within this range.
- This condition can be satisfied by certain integers, thus it is possible to have integral roots. **This option is not necessarily true.**

#### Option 4: None of these
Since we found that the third option is not necessarily true, this option is also not correct.

### Conclusion
The statement that is **not true** is:
- **Option 3: Cannot have an integral root if \( n^2 - 1 < a < n^2 + 2n \)**.