If `a gt b gt 0` are two real numbers, the value of, `sqrt(ab+(a-b)sqrt(ab+(a-b)sqrt(ab+(a-b)sqrt(ab+….))))` is

Options

A. independent of b
B. independent of a
C. independent of both a & b
D. dependent of both a & b

Explanation

To solve the problem, we need to evaluate the expression:

\[ x = \sqrt{ab + (a - b) \sqrt{ab + (a - b) \sqrt{ab + (a - b) \sqrt{ab + \ldots}}}} \]

Let's denote the entire expression under the square root as \( x \). Therefore, we can rewrite the equation as:

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

Now, we will square both sides to eliminate the square root:

\[ x^2 = ab + (a - b)x \]

Rearranging this equation gives us:

\[ x^2 - (a - b)x - ab = 0 \]

This is a quadratic equation in \( x \). We can solve it using the quadratic formula:

\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]

where \( A = 1 \), \( B = -(a - b) \), and \( C = -ab \).

Substituting these values into the formula gives:

\[ x = \frac{(a - b) \pm \sqrt{(a - b)^2 - 4 \cdot 1 \cdot (-ab)}}{2 \cdot 1} \]

This simplifies to:

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

Now, let's simplify the term under the square root:

\[ (a - b)^2 + 4ab = a^2 - 2ab + b^2 + 4ab = a^2 + 2ab + b^2 = (a + b)^2 \]

Thus, we have:

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

Now, we consider the two cases for the \( \pm \):

1. **Case 1**: Using the positive sign:
\[ x = \frac{(a - b) + (a + b)}{2} = \frac{2a}{2} = a \]

2. **Case 2**: Using the negative sign:
\[ x = \frac{(a - b) - (a + b)}{2} = \frac{-2b}{2} = -b \]

Since \( a > b > 0 \), we discard the negative solution \( -b \) because \( x \) must be positive. Therefore, we conclude that:

\[ x = a \]

Thus, the value of the expression is:

\[ \boxed{a} \]