Will give 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. ax+b+c=d 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.

Part 1: We will create the two equations using the model a√(x + b) + c = d,

where one of them has an extraneous solution while the other one has a non-extraneous solution.

The equation for an extraneous solution would be:

√(x+ 2) + 3 = 0

and for the non extraneous solution will be:

√(x + 2) + 3 = 6

Part 2: Solving the equation for the extraneous solution:

√(x + 2) + 3 = 0

√(x + 2) = - 3     

Squaring both the sides to get:

x + 2 = 9

x = 7

and solving the equation for the non-extraneous solution:

√(x + 2) + 3 = 6

√(x + 2) = 3 

Squaring both the sides to get:

x + 2 = 9

x = 7

Part 3:  The first equation has an extraneous solution since it does not satisfy the equation if you substitute a value of 7 in it.

Proof:

√(x + 2) + 3 = 0

when x = 7

√(7 + 2) + 3 = 0

√(9) + 3 = 0

3+ 3 = 0

6= 0

Hence it is false.

step-by-step explanation: