To find the value of $x$ in the equation $\sqrt{x+1} + \sqrt{x-5} = 8$, we will isolate one radical, square both sides, and then repeat the process. Step 1: Isolate one of the square root terms. Subtract $\sqrt{x-5}$ from both sides: $$ \sqrt{x+1} = 8 - \sqrt{x-5} $$ Step 2: Square both sides of the equation. $$ (\sqrt{x+1})^2 = (8 - \sqrt{x-5})^2 $$ $$ x+1 = 8^2 - 2(8)\sqrt{x-5} + (\sqrt{x-5})^2 $$ $$ x+1 = 64 - 16\sqrt{x-5} + (x-5) $$ Step 3: Simplify the equation. $$ x+1 = x + 59 - 16\sqrt{x-5} $$ Step 4: Isolate the remaining square root term. Subtract $x$ from both sides: $$ 1 = 59 - 16\sqrt{x-5} $$ Subtract 59 from both sides: $$ 1 - 59 = -16\sqrt{x-5} $$ $$ -58 = -16\sqrt{x-5} $$ Divide by -16: $$ \frac{-58}{-16} = \sqrt{x-5} $$ $$ \frac{29}{8} = \sqrt{x-5} $$ Step 5: Square both sides again. $$ \left(\frac{29}{8}\right)^2 = (\sqrt{x-5})^2 $$ $$ \frac{29^2}{8^2} = x-5 $$ $$ \frac{841}{64} = x-5 $$ Step 6: Solve for $x$. Add 5 to both sides: $$ x = \frac{841}{64} + 5 $$ To add, find a common denominator: $$ x = \frac{841}{64} + \frac{5 \times 64}{64} $$ $$ x = \frac{841}{64} + \frac{320}{64} $$ $$ x = \frac{841 + 320}{64} $$ $$ x = \frac{1161}{64} $$ Step 7: Check the solution. Substitute $x = \frac{1161}{64}$ into the original equation: $$ \sqrt{\frac{1161}{64}+1} + \sqrt{\frac{1161}{64}-5} = 8 $$ $$ \sqrt{\frac{1161+64}{64}} + \sqrt{\frac{1161-320}{64}} = 8 $$ $$ \sqrt{\frac{1225}{64}} + \sqrt{\frac{841}{64}} = 8 $$ $$ \frac{\sqrt{1225}}{\sqrt{64}} + \frac{\sqrt{841}}{\sqrt{64}} = 8 $$ $$ \frac{35}{8} + \frac{29}{8} = 8 $$ $$ \frac{35+29}{8} = 8 $$ $$ \frac{64}{8} = 8 $$ $$ 8 = 8 $$ The solution is correct. The value of $x$ is $\boxed{\frac{1161}{64}}$.