`sqrt(t+4)=t-2`
Which of the following contains the solution set to the equation above?

Options

A. `{0, 5}`
B. `{0, 4, 5}`
C. `{0}`
D. `{5}`

Answer

To solve the equation \( \sqrt{t+4} = t - 2 \), we will follow these steps:

### Step 1: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ (\sqrt{t+4})^2 = (t - 2)^2 \]
This simplifies to:
\[ t + 4 = (t - 2)(t - 2) \]

### Step 2: Expand the right side
Now we expand the right side using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \):
\[ t + 4 = t^2 - 4t + 4 \]

### Step 3: Rearrange the equation
Next, we rearrange the equation to bring all terms to one side:
\[ t + 4 - 4 = t^2 - 4t \]
This simplifies to:
\[ t + 4 - 4 = t^2 - 4t \implies t = t^2 - 4t \]
Rearranging gives:
\[ 0 = t^2 - 4t - t \]
Thus, we have:
\[ 0 = t^2 - 5t \]

### Step 4: Factor the equation
Now we can factor the quadratic equation:
\[ 0 = t(t - 5) \]

### Step 5: Solve for \( t \)
Setting each factor to zero gives us the solutions:
\[ t = 0 \quad \text{or} \quad t - 5 = 0 \implies t = 5 \]

### Step 6: Check for extraneous solutions
We need to check if these solutions satisfy the original equation:
1. For \( t = 0 \):
\[ \sqrt{0 + 4} = 0 - 2 \implies \sqrt{4} = -2 \quad \text{(not valid)} \]
2. For \( t = 5 \):
\[ \sqrt{5 + 4} = 5 - 2 \implies \sqrt{9} = 3 \quad \text{(valid)} \]

### Conclusion
The only valid solution is \( t = 5 \). Therefore, the solution set to the equation is:
\[ \{5\} \]