1. We start with the equation \(\sqrt{8 - x} = 2x + d\) and want to find \(d\) such that \(x = -1\) is an extraneous solution. 2. An extraneous solution is a root of the squared equation that does not satisfy the original equation. 3. Substitute \(x = -1\) into the right side: $$2(-1) + d = -2 + d$$ 4. Substitute \(x = -1\) into the left side: $$\sqrt{8 - (-1)} = \sqrt{9} = 3$$ 5. For \(x = -1\) to be a solution of the squared equation but not the original, \(x = -1\) must satisfy $$\left(\sqrt{8 - x}\right)^2 = (2x + d)^2$$ but $$\sqrt{8 - (-1)} \neq 2(-1) + d$$ 6. Squaring the original equation: $$8 - x = (2x + d)^2$$ Substitute \(x = -1\): $$8 - (-1) = (2(-1) + d)^2$$ $$9 = (-2 + d)^2$$ 7. Solve for \(d\): $$(-2 + d)^2 = 9$$ Take the square root: $$-2 + d = \pm 3$$ Two cases: Case 1: \(-2 + d = 3 \Rightarrow d = 5\) Case 2: \(-2 + d = -3 \Rightarrow d = -1\) 8. Check which value gives an extraneous solution by plugging into the original equation: For \(d = 5\): Right side at \(x = -1\): \(-2 + 5 = 3\) Compare to left side: \(3 = 3\), so \(x = -1\) is a valid solution, not extraneous. For \(d = -1\): Right side at \(x = -1\): \(-2 - 1 = -3\) Compare to left side: \(3 \neq -3\), so the original equation is not satisfied, making \(x = -1\) extraneous. **Final answer:** $$d = -1$$