If ` a b = 21 and ((a + b ) ^(2))/( (a - b ) ^(2)) = (25)/(4)` ten the value of `a ^(2) + b ^(2) + 3 ab` is

Options

A. 115
B. 121
C. 125
D. 127

Answer

To solve the problem step by step, we start with the given equations:

1. \( ab = 21 \)
2. \( \frac{(a + b)^2}{(a - b)^2} = \frac{25}{4} \)

### Step 1: Rewrite the second equation
From the second equation, we can take the square root of both sides:
\[ \frac{a + b}{a - b} = \frac{5}{2} \quad \text{or} \quad \frac{a + b}{a - b} = -\frac{5}{2} \]

### Step 2: Solve for \( a + b \) and \( a - b \)
Let's consider the positive case first:
\[ a + b = \frac{5}{2}(a - b) \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 2(a + b) = 5(a - b) \]
Expanding both sides:
\[ 2a + 2b = 5a - 5b \]
Rearranging gives:
\[ 2a + 2b + 5b = 5a \]
\[ 2a + 7b = 5a \]
\[ 3a = 7b \quad \Rightarrow \quad \frac{a}{b} = \frac{7}{3} \]

### Step 3: Substitute \( a \) in terms of \( b \)
From \( \frac{a}{b} = \frac{7}{3} \), we can express \( a \) as:
\[ a = \frac{7}{3}b \]

### Step 4: Substitute into \( ab = 21 \)
Now substitute \( a \) into the equation \( ab = 21 \):
\[ \left(\frac{7}{3}b\right)b = 21 \]
\[ \frac{7}{3}b^2 = 21 \]
Multiplying both sides by 3:
\[ 7b^2 = 63 \]
Dividing by 7:
\[ b^2 = 9 \quad \Rightarrow \quad b = 3 \]

### Step 5: Find \( a \)
Now substitute \( b = 3 \) back into \( a = \frac{7}{3}b \):
\[ a = \frac{7}{3} \cdot 3 = 7 \]

### Step 6: Calculate \( a^2 + b^2 + 3ab \)
Now we can calculate \( a^2 + b^2 + 3ab \):
\[ a^2 = 7^2 = 49 \]
\[ b^2 = 3^2 = 9 \]
\[ 3ab = 3 \cdot 7 \cdot 3 = 63 \]
Adding these together:
\[ a^2 + b^2 + 3ab = 49 + 9 + 63 = 121 \]

### Final Answer
Thus, the value of \( a^2 + b^2 + 3ab \) is \( \boxed{121} \).