HELP PLEASE! I'll be your fan

2 answers:

6 0

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

Irina-Kira [14]3 years ago

3 0

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

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