In each of the following find the value of 'a'
`9x^(2) + (7a-5)x + 25 = (3x + 5)^(2)`

Answer

To solve the equation \( 9x^{2} + (7a-5)x + 25 = (3x + 5)^{2} \) and find the value of \( a \), we can follow these steps:

### Step 1: Expand the right-hand side
We start by expanding the right-hand side of the equation using the formula for the square of a binomial:
\[ (a + b)^{2} = a^{2} + 2ab + b^{2} \]
Here, \( a = 3x \) and \( b = 5 \). Therefore:
\[ (3x + 5)^{2} = (3x)^{2} + 2(3x)(5) + (5)^{2} \]
Calculating this gives:
\[ (3x + 5)^{2} = 9x^{2} + 30x + 25 \]

### Step 2: Set the left-hand side equal to the expanded right-hand side
Now we can rewrite the equation:
\[ 9x^{2} + (7a - 5)x + 25 = 9x^{2} + 30x + 25 \]

### Step 3: Cancel out the common terms
Next, we can cancel \( 9x^{2} \) and \( 25 \) from both sides:
\[ (7a - 5)x = 30x \]

### Step 4: Equate the coefficients of \( x \)
Since the coefficients of \( x \) must be equal, we can set:
\[ 7a - 5 = 30 \]

### Step 5: Solve for \( a \)
Now, we solve for \( a \):
\[ 7a = 30 + 5 \]
\[ 7a = 35 \]
\[ a = \frac{35}{7} \]
\[ a = 5 \]

### Final Answer
Thus, the value of \( a \) is \( 5 \).

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