If `x=(sqrt(a+2b)+sqrt(a-2b))/(sqrt(a+2b)-sqrt(a-2b))`, then find the value of `bx^(2)-ax+b`.

Options

A. `2`
B. `1`
C. `0`
D. `6`

Answer

To solve the problem, we start with the given expression for \( x \):

\[ x = \frac{\sqrt{a + 2b} + \sqrt{a - 2b}}{\sqrt{a + 2b} - \sqrt{a - 2b}} \]

We need to find the value of \( bx^2 - ax + b \).

### Step 1: Simplifying \( x \)

To simplify \( x \), we can multiply the numerator and denominator by the conjugate of the denominator:

\[ x = \frac{(\sqrt{a + 2b} + \sqrt{a - 2b})(\sqrt{a + 2b} + \sqrt{a - 2b})}{(\sqrt{a + 2b} - \sqrt{a - 2b})(\sqrt{a + 2b} + \sqrt{a - 2b})} \]

The denominator simplifies using the difference of squares:

\[ (\sqrt{a + 2b})^2 - (\sqrt{a - 2b})^2 = (a + 2b) - (a - 2b) = 4b \]

The numerator becomes:

\[ (\sqrt{a + 2b} + \sqrt{a - 2b})^2 = (a + 2b) + (a - 2b) + 2\sqrt{(a + 2b)(a - 2b)} = 2a + 2\sqrt{(a + 2b)(a - 2b)} \]

Thus, we have:

\[ x = \frac{2a + 2\sqrt{(a + 2b)(a - 2b)}}{4b} = \frac{a + \sqrt{(a + 2b)(a - 2b)}}{2b} \]

### Step 2: Finding \( x^2 \)

Next, we compute \( x^2 \):

\[ x^2 = \left(\frac{a + \sqrt{(a + 2b)(a - 2b)}}{2b}\right)^2 = \frac{(a + \sqrt{(a + 2b)(a - 2b)})^2}{4b^2} \]

Expanding the numerator:

\[ (a + \sqrt{(a + 2b)(a - 2b)})^2 = a^2 + 2a\sqrt{(a + 2b)(a - 2b)} + (a + 2b)(a - 2b) \]

The term \( (a + 2b)(a - 2b) \) simplifies to \( a^2 - 4b^2 \). Thus:

\[ x^2 = \frac{a^2 + 2a\sqrt{(a + 2b)(a - 2b)} + a^2 - 4b^2}{4b^2} = \frac{2a^2 - 4b^2 + 2a\sqrt{(a + 2b)(a - 2b)}}{4b^2} \]

### Step 3: Finding \( bx^2 - ax + b \)

Now we substitute \( x \) and \( x^2 \) into \( bx^2 - ax + b \):

\[ bx^2 = b\left(\frac{2a^2 - 4b^2 + 2a\sqrt{(a + 2b)(a - 2b)}}{4b^2}\right) = \frac{2a^2 - 4b^2 + 2a\sqrt{(a + 2b)(a - 2b)}}{4b} \]

Now substituting into \( bx^2 - ax + b \):

\[ bx^2 - ax + b = \frac{2a^2 - 4b^2 + 2a\sqrt{(a + 2b)(a - 2b)}}{4b} - a\left(\frac{a + \sqrt{(a + 2b)(a - 2b)}}{2b}\right) + b \]

Combining these terms leads to:

\[ = 0 \]

Thus, the final result is:

\[ bx^2 - ax + b = 0 \]

### Final Answer

The value of \( bx^2 - ax + b \) is \( 0 \).