If `c ne 0` and the equation `(p)/(2x)=(a)/(x+c)+(b)/(x-c)` has two equal roots, then `p` can be

Options

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

Explanation

To solve the equation \(\frac{p}{2x} = \frac{a}{x+c} + \frac{b}{x-c}\) for the condition that it has two equal roots, we will follow these steps:

### Step 1: Combine the Right Side of the Equation
We start by finding a common denominator for the right-hand side:

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

This simplifies to:

\[ \frac{p}{2x} = \frac{ax - ac + bx + bc}{x^2 - c^2} \]

Combining the terms in the numerator gives:

\[ \frac{p}{2x} = \frac{(a+b)x + (bc - ac)}{x^2 - c^2} \]

### Step 2: Cross Multiply
Cross-multiplying gives us:

\[ p(x^2 - c^2) = 2x((a+b)x + (bc - ac)) \]

Expanding both sides results in:

\[ px^2 - pc^2 = 2(a+b)x^2 + 2(bc - ac)x \]

### Step 3: Rearranging the Equation
Rearranging the equation leads to:

\[ (px^2 - 2(a+b)x^2 - 2(bc - ac)x - pc^2 = 0 \]

This can be simplified to:

\[ (p - 2(a+b))x^2 - 2(bc - ac)x - pc^2 = 0 \]

### Step 4: Condition for Equal Roots
For the quadratic equation \(Ax^2 + Bx + C = 0\) to have equal roots, the discriminant must be zero. The discriminant \(D\) is given by:

\[ D = B^2 - 4AC \]

In our case:

- \(A = p - 2(a+b)\)
- \(B = -2(bc - ac)\)
- \(C = -pc^2\)

Setting the discriminant to zero gives:

\[ (-2(bc - ac))^2 - 4(p - 2(a+b))(-pc^2) = 0 \]

### Step 5: Simplifying the Discriminant
Expanding this leads to:

\[ 4(bc - ac)^2 + 4pc^2(p - 2(a+b)) = 0 \]

Dividing through by 4 simplifies to:

\[ (bc - ac)^2 + pc^2(p - 2(a+b)) = 0 \]

### Step 6: Solving for \(p\)
Since \(c \neq 0\), we can isolate \(p\):

\[ pc^2(p - 2(a+b)) = -(bc - ac)^2 \]

This implies that \(p\) can take on certain values based on the parameters \(a\) and \(b\).

### Conclusion
After analyzing the discriminant and the conditions for equal roots, we find that \(p\) can be expressed in terms of \(a\) and \(b\) as follows:

\[ p = a + b \pm 2\sqrt{ab} \]

Thus, the possible values for \(p\) can be:

1. \(\sqrt{A} - \sqrt{B}\) whole square
2. \(\sqrt{A} + \sqrt{B}\) whole square
3. \(A + B\)
4. \(A - B\)