\[ \sqrt{b+27}+3=b \] The solution set is

To solve the equation \( \sqrt{b+27} + 3 = b \), we can first isolate the square root: \[ \sqrt{b+27} = b - 3 \] Next, square both sides of the equation: \[ b + 27 = (b - 3)^2 \] Expanding the right side: \[ b + 27 = b^2 - 6b + 9 \] Rearranging the equation gives: \[ 0 = b^2 - 7b - 18 \] Now, we can factor or use the quadratic formula to find \( b \). Factoring gives: \[ 0 = (b - 9)(b + 2) \] Setting each factor to zero results in: \[ b - 9 = 0 \quad \Rightarrow \quad b = 9 \] \[ b + 2 = 0 \quad \Rightarrow \quad b = -2 \] We must check both solutions in the original equation to ensure they are valid. For \( b = 9 \): \[ \sqrt{9 + 27} + 3 = \sqrt{36} + 3 = 6 + 3 = 9 \quad \text{(valid)} \] For \( b = -2 \): \[ \sqrt{-2 + 27} + 3 = \sqrt{25} + 3 = 5 + 3 = 8 \quad \text{(not valid)} \] Thus, the solution set is \( \{9\} \).