The value of `b` for which the equation `2(log)_(1/(25))(bx+28)=-(log)_5(12-4x-x^2)` has coincident roots is

Options

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

Explanation

To solve the equation \(2 \log_{1/25}(bx + 28) = -\log_{5}(12 - 4x - x^2)\) for the value of \(b\) that results in coincident roots, we will follow these steps:

### Step 1: Simplify the Left-Hand Side (LHS)
The LHS is given as:
\[ 2 \log_{1/25}(bx + 28) \]
We can rewrite \(1/25\) as \(5^{-2}\), so:
\[ \log_{1/25}(bx + 28) = \log_{5^{-2}}(bx + 28) = -\frac{1}{2} \log_{5}(bx + 28) \]
Thus, we have:
\[ 2 \log_{1/25}(bx + 28) = 2 \left(-\frac{1}{2} \log_{5}(bx + 28)\right) = -\log_{5}(bx + 28) \]

### Step 2: Set the LHS Equal to the Right-Hand Side (RHS)
Now we can set the simplified LHS equal to the RHS:
\[ -\log_{5}(bx + 28) = -\log_{5}(12 - 4x - x^2) \]
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 equate the arguments:
\[ bx + 28 = 12 - 4x - x^2 \]

### Step 4: Rearrange the Equation
Rearranging gives:
\[ x^2 + (b + 4)x + (28 - 12) = 0 \]
This simplifies to:
\[ x^2 + (b + 4)x + 16 = 0 \]

### Step 5: Condition for Coincident Roots
For the quadratic equation \(ax^2 + bx + c = 0\) to have coincident roots, the discriminant must be zero:
\[ D = B^2 - 4AC = 0 \]
Here, \(A = 1\), \(B = b + 4\), and \(C = 16\). Thus:
\[ (b + 4)^2 - 4 \cdot 1 \cdot 16 = 0 \]

### Step 6: Solve the Discriminant Equation
Expanding the discriminant:
\[ (b + 4)^2 - 64 = 0 \]
This leads to:
\[ (b + 4)^2 = 64 \]
Taking the square root of both sides gives:
\[ b + 4 = 8 \quad \text{or} \quad b + 4 = -8 \]
Solving these equations:
1. \(b + 4 = 8 \implies b = 4\)
2. \(b + 4 = -8 \implies b = -12\)

### Final Answer
Thus, the values of \(b\) for which the equation has coincident roots are:
\[ b = 4 \quad \text{or} \quad b = -12 \]