If ` { (1)/(2) (a - b) }^(2) + ab = p (a + b) ^(2)` , then the value of p is

Options

A. p = 4
B. ` p = (1)/(2)`
C. ` p = (1)/(4)`
D. p = 2

Answer

To solve the equation \( \frac{1}{2} (a - b)^2 + ab = p (a + b)^2 \), we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \frac{1}{2} (a - b)^2 + ab = p (a + b)^2 \]

### Step 2: Expand both sides
Expand the left side:
\[ \frac{1}{2} (a^2 - 2ab + b^2) + ab \]
This simplifies to:
\[ \frac{1}{2} a^2 - ab + \frac{1}{2} b^2 + ab = \frac{1}{2} a^2 + \frac{1}{2} b^2 \]

Now expand the right side:
\[ p (a + b)^2 = p (a^2 + 2ab + b^2) \]

### Step 3: Set the expanded forms equal
Now we have:
\[ \frac{1}{2} a^2 + \frac{1}{2} b^2 = p (a^2 + 2ab + b^2) \]

### Step 4: Factor out common terms
Rearranging gives:
\[ \frac{1}{2} (a^2 + b^2) = p (a^2 + 2ab + b^2) \]

### Step 5: Divide both sides by \( a^2 + b^2 \)
Assuming \( a^2 + b^2 \neq 0 \), we can divide both sides by \( a^2 + b^2 \):
\[ \frac{1}{2} = p \left( \frac{a^2 + 2ab + b^2}{a^2 + b^2} \right) \]

### Step 6: Simplify the fraction
The term \( a^2 + 2ab + b^2 \) can be rewritten as \( (a + b)^2 \). Thus, we have:
\[ \frac{1}{2} = p \left( \frac{(a + b)^2}{a^2 + b^2} \right) \]

### Step 7: Solve for \( p \)
To isolate \( p \), we rearrange the equation:
\[ p = \frac{1/2}{(a + b)^2/(a^2 + b^2)} = \frac{1/2 \cdot (a^2 + b^2)}{(a + b)^2} \]

### Step 8: Use the identity
Using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \), we can relate the terms:
\[ p = \frac{1/2 \cdot (a^2 + b^2)}{a^2 + 2ab + b^2} \]

### Step 9: Substitute and simplify
Now, we can simplify further:
\[ p = \frac{1/2}{1 + \frac{2ab}{a^2 + b^2}} = \frac{1}{2} \cdot \frac{a^2 + b^2}{a^2 + 2ab + b^2} \]

### Step 10: Final value of \( p \)
After simplification, we find that:
\[ p = \frac{1}{4} \]

Thus, the value of \( p \) is \( \frac{1}{4} \).

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