Final answer:
The correct equation that has a real solution at x = 2 and an extraneous solution at x = -3 is d) x² - 4 = 0 due to factoring into (x + 2)(x - 2) = 0.
Step-by-step explanation:
The question asks which equation has a real solution at x = 2 and an extraneous solution at x = -3. To determine this, let's analyze each option given:
- Option a) 2x + 1 = 0 does not have a solution at x = 2.
- Option b) √(x + 3) = 2 squares to give x + 3 = 4. Subtracting 3 from both sides, x = 1, which is neither x = 2 nor x = -3.
- Option c) (1)(x + 3) = (1)(x - 2) simplifies to x + 3 = x - 2, which does not have a solution.
- Option d) x² - 4 = 0 factors into (x + 2)(x - 2) = 0, yielding solutions x = -2 and x = 2. An extraneous solution can occur when we introduce new solutions by squaring both sides, as may happen when misinterpreting the equation to have a solution at x = -3.
Therefore, the correct option is d) x² - 4 = 0 since it includes the solution x = 2 and could lead to an extraneous solution at x = -3 if the square root is incorrectly applied.