If both roots of the equation `x^2-2ax + a^2-1=0` lie between `(-2,2)` then a lies in the interval

Options

A. `R`
B. `(-1,1)`
C. `(-2,2)`
D. `(-3,-1)uu(1,3)`

Explanation

To solve the problem, we need to determine the interval for \( a \) such that both roots of the equation \( x^2 - 2ax + (a^2 - 1) = 0 \) lie between \(-2\) and \(2\).

### Step 1: Find the roots of the 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 \( x^2 - 2ax + (a^2 - 1) = 0 \):
- \( a = 1 \)
- \( b = -2a \)
- \( c = a^2 - 1 \)

Substituting these values into the quadratic formula gives:
\[ x = \frac{2a \pm \sqrt{(-2a)^2 - 4 \cdot 1 \cdot (a^2 - 1)}}{2 \cdot 1} \]
\[ x = \frac{2a \pm \sqrt{4a^2 - 4(a^2 - 1)}}{2} \]
\[ x = \frac{2a \pm \sqrt{4a^2 - 4a^2 + 4}}{2} \]
\[ x = \frac{2a \pm \sqrt{4}}{2} \]
\[ x = \frac{2a \pm 2}{2} \]
\[ x = a \pm 1 \]

Thus, the roots are \( x_1 = a + 1 \) and \( x_2 = a - 1 \).

### Step 2: Set the conditions for the roots to lie between -2 and 2
We need both roots \( x_1 \) and \( x_2 \) to lie within the interval \((-2, 2)\).

1. For the first root \( x_1 = a + 1 \):
\[ -2 < a + 1 < 2 \]
This can be split into two inequalities:
- \( -2 < a + 1 \) leads to \( a > -3 \)
- \( a + 1 < 2 \) leads to \( a < 1 \)

Therefore, from the first root, we have:
\[ -3 < a < 1 \]

2. For the second root \( x_2 = a - 1 \):
\[ -2 < a - 1 < 2 \]
This can also be split into two inequalities:
- \( -2 < a - 1 \) leads to \( a > -1 \)
- \( a - 1 < 2 \) leads to \( a < 3 \)

Therefore, from the second root, we have:
\[ -1 < a < 3 \]

### Step 3: Combine the intervals
Now we combine the intervals obtained from both roots:
1. From \( x_1 \): \( -3 < a < 1 \)
2. From \( x_2 \): \( -1 < a < 3 \)

The common interval where both conditions are satisfied is:
\[ -1 < a < 1 \]

### Final Answer
Thus, the value of \( a \) lies in the interval:
\[ (-1, 1) \]