IF ` alpha , beta ` are real and ` alpha ^2 , - beta ^2` are the roots of ` a ^2 x^2+x+1-a^2=0 (A gt 1)` then ` beta ^2` =

Options

A. `a^(2)`
B. 1
C. `1-a^(2)`
D. `1+a^(2)`

Answer

To solve the problem, we need to find the value of \( \beta^2 \) given that \( \alpha^2 \) and \( -\beta^2 \) are the roots of the quadratic equation \( a^2 x^2 + x + 1 - a^2 = 0 \) where \( a > 1 \).

### Step-by-Step Solution:

1. **Identify the coefficients of the quadratic equation:**
The equation can be rewritten as:
\[ a^2 x^2 + x + (1 - a^2) = 0 \]
Here, \( A = a^2 \), \( B = 1 \), and \( C = 1 - a^2 \).

2. **Use the properties of roots:**
For a quadratic equation \( Ax^2 + Bx + C = 0 \), the sum of the roots \( r_1 + r_2 = -\frac{B}{A} \) and the product of the roots \( r_1 \cdot r_2 = \frac{C}{A} \).
Let the roots be \( r_1 = \alpha^2 \) and \( r_2 = -\beta^2 \).

3. **Calculate the sum of the roots:**
\[ \alpha^2 - \beta^2 = -\frac{1}{a^2} \]

4. **Calculate the product of the roots:**
\[ \alpha^2 \cdot (-\beta^2) = \frac{1 - a^2}{a^2} \]
This simplifies to:
\[ -\alpha^2 \beta^2 = \frac{1 - a^2}{a^2} \]

5. **Express \( \alpha^2 \) in terms of \( \beta^2 \):**
From the sum of the roots equation, we can express \( \alpha^2 \):
\[ \alpha^2 = -\frac{1}{a^2} + \beta^2 \]

6. **Substitute \( \alpha^2 \) into the product of the roots equation:**
Substitute \( \alpha^2 \) into the product equation:
\[ -\left(-\frac{1}{a^2} + \beta^2\right) \beta^2 = \frac{1 - a^2}{a^2} \]
This simplifies to:
\[ \left(\frac{1}{a^2} - \beta^2\right) \beta^2 = \frac{1 - a^2}{a^2} \]

7. **Multiply through by \( a^2 \) to eliminate the denominator:**
\[ (1 - a^2 \beta^2) \beta^2 = 1 - a^2 \]

8. **Rearranging gives us a quadratic in \( \beta^2 \):**
\[ a^2 \beta^4 - \beta^2 + (1 - a^2) = 0 \]

9. **Use the quadratic formula to solve for \( \beta^2 \):**
The quadratic formula states \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here \( a = a^2 \), \( b = -1 \), and \( c = 1 - a^2 \):
\[ \beta^2 = \frac{1 \pm \sqrt{(-1)^2 - 4(a^2)(1 - a^2)}}{2a^2} \]
Simplifying the discriminant:
\[ 1 - 4a^2 + 4a^4 = 4a^4 - 4a^2 + 1 = (2a^2 - 1)^2 \]
Thus,
\[ \beta^2 = \frac{1 \pm (2a^2 - 1)}{2a^2} \]

10. **Find the two possible values for \( \beta^2 \):**
- First case:
\[ \beta^2 = \frac{2a^2}{2a^2} = 1 \]
- Second case:
\[ \beta^2 = \frac{1 - (2a^2 - 1)}{2a^2} = \frac{2 - 2a^2}{2a^2} = \frac{1 - a^2}{a^2} \]
Since \( a > 1 \), \( 1 - a^2 < 0 \), hence this case yields a negative value which is not possible for \( \beta^2 \).

### Conclusion:
Thus, the only valid solution is:
\[ \beta^2 = 1 \]