If `alpha, beta` are real and `alpha^(2), beta^(2)` are the roots of the equation `a^(2)x^(2)-x+1-a^(2)=0 (1/sqrt2 lt a lt 1) and beta^(2) ne 1," then "beta^(2)=`

Options

A. `a^(2)`
B. `(1-a^(2))/(a^(2))`
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 \( \frac{1}{\sqrt{2}} < a < 1 \) and \( \beta^2 \neq 1 \).

### Step 1: Identify the coefficients of the quadratic equation
The quadratic equation can be expressed in the standard form \( Ax^2 + Bx + C = 0 \) where:
- \( A = a^2 \)
- \( B = -1 \)
- \( C = 1 - a^2 \)

### Step 2: Use the sum and product of roots
For a quadratic equation \( Ax^2 + Bx + C = 0 \):
- The sum of the roots \( \alpha^2 + \beta^2 = -\frac{B}{A} = \frac{1}{a^2} \)
- The product of the roots \( \alpha^2 \beta^2 = \frac{C}{A} = \frac{1 - a^2}{a^2} \)

### Step 3: Express \( \alpha^2 \) in terms of \( \beta^2 \)
From the sum of roots:
\[ \alpha^2 + \beta^2 = \frac{1}{a^2} \]
We can express \( \alpha^2 \) as:
\[ \alpha^2 = \frac{1}{a^2} - \beta^2 \]

### Step 4: Substitute \( \alpha^2 \) into the product of roots
Now substitute \( \alpha^2 \) into the product of roots:
\[ \left(\frac{1}{a^2} - \beta^2\right) \beta^2 = \frac{1 - a^2}{a^2} \]

### Step 5: Expand and rearrange the equation
Expanding the left-hand side gives:
\[ \frac{\beta^2}{a^2} - \beta^4 = \frac{1 - a^2}{a^2} \]
Multiplying through by \( a^2 \) to eliminate the denominator:
\[ \beta^2 - a^2 \beta^4 = 1 - a^2 \]
Rearranging this gives:
\[ a^2 \beta^4 - \beta^2 + (1 - a^2) = 0 \]

### Step 6: Let \( y = \beta^2 \) and rewrite the equation
Let \( y = \beta^2 \):
\[ a^2 y^2 - y + (1 - a^2) = 0 \]

### Step 7: Apply the quadratic formula
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = a^2 \), \( b = -1 \), and \( c = 1 - a^2 \):
\[ y = \frac{1 \pm \sqrt{(-1)^2 - 4(a^2)(1 - a^2)}}{2a^2} \]
\[ y = \frac{1 \pm \sqrt{1 - 4a^2 + 4a^4}}{2a^2} \]
\[ y = \frac{1 \pm \sqrt{(2a^2 - 1)^2}}{2a^2} \]

### Step 8: Simplify the expression
Since \( \sqrt{(2a^2 - 1)^2} = |2a^2 - 1| \):
Given \( \frac{1}{\sqrt{2}} < a < 1 \), we have \( 2a^2 - 1 > 0 \), thus:
\[ y = \frac{1 + (2a^2 - 1)}{2a^2} \quad \text{or} \quad y = \frac{1 - (2a^2 - 1)}{2a^2} \]
This simplifies to:
\[ y = \frac{2a^2}{2a^2} = 1 \quad \text{or} \quad y = \frac{2 - 2a^2}{2a^2} = \frac{1 - a^2}{a^2} \]

### Step 9: Determine the valid solution for \( \beta^2 \)
Since \( \beta^2 \neq 1 \), we take:
\[ \beta^2 = \frac{1 - a^2}{a^2} \]

### Final Answer
Thus, the value of \( \beta^2 \) is:
\[ \beta^2 = \frac{1 - a^2}{a^2} \]