If `rlt0 and (4r-4)^(2)=36`, what is the value of r?

Options

A. `-6`
B. `-2`
C. `-1`
D. `-(1)/(2)`

Explanation

To solve the equation \( (4r - 4)^2 = 36 \) under the condition that \( r < 0 \), we will follow these steps:

### Step 1: Take the square root of both sides
We start with the equation:
\[ (4r - 4)^2 = 36 \]
Taking the square root of both sides gives us:
\[ 4r - 4 = \pm 6 \]

### Step 2: Solve for \( 4r \)
This gives us two separate equations to solve:
1. \( 4r - 4 = 6 \)
2. \( 4r - 4 = -6 \)

### Step 3: Solve the first equation
For the first equation:
\[ 4r - 4 = 6 \]
Adding 4 to both sides:
\[ 4r = 10 \]
Dividing by 4:
\[ r = \frac{10}{4} = \frac{5}{2} \]

### Step 4: Solve the second equation
For the second equation:
\[ 4r - 4 = -6 \]
Adding 4 to both sides:
\[ 4r = -2 \]
Dividing by 4:
\[ r = \frac{-2}{4} = -\frac{1}{2} \]

### Step 5: Determine the valid solution
We have two potential solutions for \( r \):
1. \( r = \frac{5}{2} \)
2. \( r = -\frac{1}{2} \)

Since the problem states that \( r < 0 \), we discard \( r = \frac{5}{2} \) and keep:
\[ r = -\frac{1}{2} \]

### Final Answer
Thus, the value of \( r \) is:
\[ \boxed{-\frac{1}{2}} \]

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