Question

HELP PLEASE! I'll be your fan

Answer 1

Part a)

for extraneous solution

1 ⋅ sqrt(x+2) + 3 = 0

for non extraneous solution

1⋅ sqrt(x+2)  + 3 = 6

part b) solve each equation

1⋅sqrt(x+2)+3=0 

x+sqrt(x+2)=−3 

square both sides

(sqrt(x+2))^ 2 = (−3)^2 


x+2=9 

x=9−2 

x=7 

do you see why its extraneous

Answer 2

Hello there.

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.

a√x+b+c=d

1⋅ sqrt(x+2)  + 3 = 6