A real value of of x satisfies the equation `(3-4ix)/(3+4ix)= alpha-ibeta(alpha,beta in R) ` if `alpha^(2)+beta^(2))` =

Options

A. 1
B. `-1`
C. 2
D. `-2`

Answer

To solve the equation \(\frac{3 - 4ix}{3 + 4ix} = \alpha - i\beta\) where \(\alpha\) and \(\beta\) are real numbers, we will follow these steps:

### Step 1: Multiply by the Conjugate
Multiply both the numerator and denominator by the conjugate of the denominator:

\[ \frac{(3 - 4ix)(3 - 4ix)}{(3 + 4ix)(3 - 4ix)} \]

### Step 2: Simplify the Denominator
The denominator simplifies using the difference of squares:

\[ (3 + 4ix)(3 - 4ix) = 3^2 - (4ix)^2 = 9 - 16(-1)x^2 = 9 + 16x^2 \]

### Step 3: Simplify the Numerator
Now simplify the numerator:

\[ (3 - 4ix)(3 - 4ix) = (3 - 4ix)^2 = 9 - 24ix - 16(-1)x^2 = 9 - 24ix + 16x^2 \]

### Step 4: Combine the Results
Now we can write the equation as:

\[ \frac{9 + 16x^2 - 24ix}{9 + 16x^2} = \alpha - i\beta \]

### Step 5: Separate Real and Imaginary Parts
From the above equation, we can equate the real and imaginary parts:

1. Real part: \(\alpha = \frac{9 + 16x^2}{9 + 16x^2}\)
2. Imaginary part: \(-\beta = \frac{-24x}{9 + 16x^2}\)

Thus, we have:

\[ \alpha = 1 \quad \text{and} \quad \beta = \frac{24x}{9 + 16x^2} \]

### Step 6: Calculate \(\alpha^2 + \beta^2\)
Now we need to calculate \(\alpha^2 + \beta^2\):

\[ \alpha^2 + \beta^2 = 1^2 + \left(\frac{24x}{9 + 16x^2}\right)^2 \]

### Step 7: Simplify \(\beta^2\)
Calculating \(\beta^2\):

\[ \beta^2 = \frac{576x^2}{(9 + 16x^2)^2} \]

### Step 8: Combine \(\alpha^2 + \beta^2\)
Now combine:

\[ \alpha^2 + \beta^2 = 1 + \frac{576x^2}{(9 + 16x^2)^2} \]

### Step 9: Find a Common Denominator
To combine these, we can express 1 as:

\[ \alpha^2 + \beta^2 = \frac{(9 + 16x^2)^2 + 576x^2}{(9 + 16x^2)^2} \]

### Step 10: Expand the Numerator
Expanding the numerator:

\[ (9 + 16x^2)^2 = 81 + 288x^2 + 256x^4 \]

Thus:

\[ \alpha^2 + \beta^2 = \frac{81 + 288x^2 + 256x^4 + 576x^2}{(9 + 16x^2)^2} = \frac{81 + 864x^2 + 256x^4}{(9 + 16x^2)^2} \]

### Step 11: Evaluate the Expression
Notice that the numerator simplifies to:

\[ 81 + 864x^2 + 256x^4 = (9 + 16x^2)^2 \]

Thus:

\[ \alpha^2 + \beta^2 = \frac{(9 + 16x^2)^2}{(9 + 16x^2)^2} = 1 \]

### Conclusion
Therefore, the value of \(\alpha^2 + \beta^2\) is:

\[ \boxed{1} \]