Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model: 


Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.  

Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous.  

Part 3. Explain why the first equation has an extraneous solution and the second does not. 

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1 Answer


Part 1

2√(x+1)+7=5 is a radical equation with an extraneous solution.

Part 2

2√(x+1)+7=5, 2√(x+1)=-2, √(x+1)=-1, square both sides: x+1=1, x=0.

But 2√(x+1)+7=9 when x=0 and 9≠5. So x=0 is an extraneous solution.

2√(x+1)+5=7 is a radical equation with solution x=0.

Part 3

The extraneous solution is a result of squaring because (-1)²=1²=1. This creates an ambiguity. But √(x+1) specifically implies the positive root. This is the reason why the second equation works, because the positive root satisfies the equation.

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