Final answer:
The constant d that makes x=-1 an extraneous solution in the equation sqrt(8-x) = 2x + d is any value except for 5, such as d=4.
Step-by-step explanation:
To determine which value for the constant d makes x=-1 an extraneous solution in the equation sqrt(8-x) = 2x + d, we'll need to find a value for d that, when substituted back into the original equation, will not allow x = -1 to be a valid solution. In other words, x=-1 should make the equation untrue.
First, examine the structure of the equation without any modifications to understand its parts. After verifying that the left side is a square root function and the right side is linear, we recognize that squaring both sides will eliminate the square root, leading to a quadratic equation.
Substituting x=-1 into the equation and simplifying, we get sqrt(8-(-1)) = 2(-1) + d which simplifies to sqrt(9) = -2 + d. Simplifying further, we have 3 = -2 + d. Solving for d, we get d = 5.
However, we're looking for a value of d that makes x=-1 extraneous. This would mean that x=-1 should lead to a contradiction when substituting back into the equation after solving the quadratic. Since we have determined that d = 5 satisfies the equation with x=-1, we need a different value for d that doesn't fulfill the equation. If we choose any d not equal to 5, say d=4, then x=-1 would not satisfy the equation, making it an extraneous solution.