Question

Create a radical equation that has at least one extraneous solution and explain why it has an extraneous solution." A) x = √(x + 5) B) 3√(x - 2) = 2 C) √(x - 4) = 2x + 1

Answer 1

The radical equation with an extraneous solution is B) 3√(x - 2) = 2. To solve it, isolate the radical term, square both sides, and simplify. However, the extraneous solution x = 22/9 does not satisfy the original equation. if we substitute x = 22/9 back into the original equation, we get: 3√(22/9 - 2) = 2.

Step-by-step explanation:

The radical equation that has at least one extraneous solution is: B) 3√(x - 2) = 2. To solve this equation, we need to isolate the radical term first. Step 1: Divide both sides by 3 to get √(x - 2) = 2/3. Step 2: Square both sides to eliminate the square root: (x - 2) = (2/3)^2 = 4/9. Step 3: Add 2 to both sides: x = 4/9 + 2 = 22/9.

However, if we substitute x = 22/9 back into the original equation, we get: 3√(22/9 - 2) = 2. If we substitute x = 22/9 back into the original equation, we get: 3√(22/9 - 2) = 2. The term inside the radical, 22/9 - 2, becomes negative and results in an imaginary number when taking the square root. Therefore, x = 22/9 is an extraneous solution that does not satisfy the original equation.

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