The values of b for which the equation `2log_(1/25)(bx+28)=1log_5(12-4x-x^2)` has coincident roots is/are

Options

A. b = - 12
B. b = 4
C. b = 4 or b =- 12
D. b =- 4 or b = 12

Explanation

To solve the equation \( 2 \log_{1/25}(bx + 28) = -\log_5(12 - 4x - x^2) \) for the values of \( b \) such that the equation has coincident roots, we can follow these steps:

### Step 1: Rewrite the logarithmic equation
We start by rewriting the logarithmic equation using properties of logarithms. The base \( \frac{1}{25} \) can be expressed as \( 25^{-1} \), which allows us to use the property of logarithms that states \( \log_a(b^m) = m \log_a(b) \).

\[ 2 \log_{1/25}(bx + 28) = 2 \log_{25^{-1}}(bx + 28) = -2 \log_{25}(bx + 28) \]

Thus, we can rewrite the equation as:

\[ -2 \log_{25}(bx + 28) = -\log_5(12 - 4x - x^2) \]

### Step 2: Convert the logarithm base
Next, we convert the logarithm from base 25 to base 5. Since \( 25 = 5^2 \), we can use the change of base formula:

\[ \log_{25}(A) = \frac{1}{2} \log_5(A) \]

Applying this to our equation gives:

\[ -2 \cdot \frac{1}{2} \log_5(bx + 28) = -\log_5(12 - 4x - x^2) \]

This simplifies to:

\[ -\log_5(bx + 28) = -\log_5(12 - 4x - x^2) \]

### Step 3: Remove the logarithm
Since the logarithms are equal, we can set the arguments equal to each other:

\[ bx + 28 = 12 - 4x - x^2 \]

### Step 4: Rearrange the equation
Rearranging the equation gives:

\[ x^2 + (b + 4)x + 16 = 0 \]

### Step 5: Determine the condition for coincident roots
For the quadratic equation \( ax^2 + bx + c = 0 \) to have coincident roots, the discriminant must be zero:

\[ D = B^2 - 4AC = 0 \]

In our case, \( A = 1 \), \( B = b + 4 \), and \( C = 16 \). Thus, we set up the equation:

\[ (b + 4)^2 - 4 \cdot 1 \cdot 16 = 0 \]

### Step 6: Solve for \( b \)
Expanding and solving the equation:

\[ (b + 4)^2 - 64 = 0 \]

This gives:

\[ (b + 4)^2 = 64 \]

Taking the square root of both sides:

\[ b + 4 = 8 \quad \text{or} \quad b + 4 = -8 \]

Solving these equations gives:

1. \( b + 4 = 8 \) leads to \( b = 4 \)
2. \( b + 4 = -8 \) leads to \( b = -12 \)

### Conclusion
The values of \( b \) for which the equation has coincident roots are:

\[ \boxed{4 \text{ and } -12} \]

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