Final answer:
Two radical equations are presented: one with the solution x = 22, which does not yield an extraneous solution when checked, and another with the extraneous solution x = 10, which does not hold true when plugged back into the equation.
Step-by-step explanation:
To craft two radical equations based on the model equation a√(x + b) + c = d, we can select different constants for a, b, c, and d.
Radical Equation Without an Extraneous Solution
Let's choose 2√(x + 3) - 4 = 6. Solving it step-by-step:
-
- Add 4 to both sides: 2√(x + 3) = 10.
-
- Divide both sides by 2: √(x + 3) = 5.
-
- Square both sides: x + 3 = 25.
-
- Subtract 3 from both sides: x = 22.
Checking the solution in the original equation confirms it's a valid solution, without any extraneous solutions.
Radical Equation With an Extraneous Solution
Here, let's use -1√(x - 6) + 2 = 0. Following the steps:
-
- Add -2 to both sides: -1√(x - 6) = -2.
-
- Divide both sides by -1: √(x - 6) = 2.
-
- Square both sides: x - 6 = 4.
-
- Add 6 to both sides: x = 10.
However, if we put x = 10 back into the original equation, we find it does not satisfy the equation, indicating that x = 10 is an extraneous solution.
It's important to remember that when solving radical equations, you must always verify your solutions by substituting them back into the original equation to determine whether they are true solutions or extraneous solutions.