To solve the equation \( \sqrt{3x - 11} + 5 = x + 2 \), we will first isolate the square root:
- Subtract 5 from both sides:
\[ \sqrt{3x - 11} = x + 2 - 5 \]
\[ \sqrt{3x - 11} = x - 3 \]
- Now, square both sides to eliminate the square root:
\[ (3x - 11) = (x - 3)^2 \]
- Expand the right side:
\[ (3x - 11) = x^2 - 6x + 9 \]
- Rearrange the equation to one side:
\[ 0 = x^2 - 6x + 9 - 3x + 11 \]
This simplifies to:
\[ 0 = x^2 - 9x + 20 \]
- Now, we can factor the quadratic:
\[ 0 = (x - 4)(x - 5) \]
- Set each factor to zero:
\[ x - 4 = 0 \quad \text{or} \quad x - 5 = 0 \]
This gives:
\[ x = 4 \quad \text{or} \quad x = 5 \]
- We need to check both possible solutions against the original equation to ensure they do not introduce extraneous solutions.
For \( x = 4 \): \[ \sqrt{3(4) - 11} + 5 = 4 + 2 \]
\[ \sqrt{12 - 11} + 5 = 6 \]
\[ 1 + 5 = 6 \quad \text{(True)} \]
For \( x = 5 \): \[ \sqrt{3(5) - 11} + 5 = 5 + 2 \]
\[ \sqrt{15 - 11} + 5 = 7 \]
\[ \sqrt{4} + 5 = 7 \]
\[ 2 + 5 = 7 \quad \text{(True)} \]
Both solutions are valid.
The final answer is \( x = 4 \) or \( x = 5 \).
So the correct response is:
x=4 or x=5