If `a in (-1,1),` then roots of the quadratic equation `(a-1)x^2+a x+sqrt(1-a^2)=0` are a. real b. imaginary c. both equal d. none of these

Options

A. real
B. imaginary
C. both equal
D. none of these

Explanation

To determine the nature of the roots of the quadratic equation \((a-1)x^2 + ax + \sqrt{1-a^2} = 0\) given that \(a \in (-1, 1)\), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is of the form \(Ax^2 + Bx + C = 0\), where:
- \(A = a - 1\)
- \(B = a\)
- \(C = \sqrt{1 - a^2}\)

### Step 2: Calculate the discriminant
The discriminant \(D\) of a quadratic equation is given by the formula:
\[ D = B^2 - 4AC \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ D = a^2 - 4(a - 1)\sqrt{1 - a^2} \]

### Step 3: Simplify the discriminant
Now we simplify the expression:
\[ D = a^2 - 4(a - 1)\sqrt{1 - a^2} \]
To analyze the discriminant, we need to check if \(D\) is greater than, less than, or equal to zero.

### Step 4: Analyze the discriminant
We know that \(a \in (-1, 1)\). We will check the sign of \(D\) in this interval.

1. **When \(a = 0\)**:
\[ D = 0^2 - 4(0 - 1)\sqrt{1 - 0^2} = 0 + 4 = 4 > 0 \]

2. **When \(a = \frac{1}{2}\)**:
\[ D = \left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2} - 1\right)\sqrt{1 - \left(\frac{1}{2}\right)^2} \]
\[ = \frac{1}{4} - 4\left(-\frac{1}{2}\right)\sqrt{\frac{3}{4}} = \frac{1}{4} + 2 \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \sqrt{3} \]
Since \(\sqrt{3} > 1\), \(D > 0\).

3. **When \(a\) approaches -1 or 1**:
- As \(a\) approaches -1, \(D\) remains positive.
- As \(a\) approaches 1, \(D\) also remains positive.

### Conclusion
Since \(D > 0\) for all \(a \in (-1, 1)\), the roots of the quadratic equation are real and unequal.

### Final Answer
The roots of the quadratic equation are **real**.