The set `A={x:x epsilonR,x^(2)=16` and `2x=16}` is equal to

Options

A. `phi`
B. `{14,3,4}`
C. `{3}`
D. `{4}`

Explanation

To solve the problem, we need to analyze the set \( A \) defined by the conditions \( x^2 = 16 \) and \( 2x = 16 \).

### Step 1: Solve the equation \( x^2 = 16 \)

To find the values of \( x \) that satisfy \( x^2 = 16 \):

\[ x^2 = 16 \]

Taking the square root of both sides, we get:

\[ x = \pm 4 \]

Thus, from this equation, we have two solutions: \( x = 4 \) and \( x = -4 \).

### Step 2: Solve the equation \( 2x = 16 \)

Next, we solve the equation \( 2x = 16 \):

\[ 2x = 16 \]

Dividing both sides by 2, we find:

\[ x = 8 \]

### Step 3: Combine the solutions into set \( A \)

Now, we combine the solutions from both equations into set \( A \):

From \( x^2 = 16 \), we have \( \{4, -4\} \).

From \( 2x = 16 \), we have \( \{8\} \).

Thus, the set \( A \) can be expressed as:

\[ A = \{4, -4, 8\} \]

### Step 4: Conclusion

The final set \( A \) is:

\[ A = \{4, -4, 8\} \]