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Murljashka [212]
3 years ago
13

HELP PLEASE! I'll be your fan

Mathematics
2 answers:
Bond [772]3 years ago
6 0
Part a)

for extraneous solution
<span>
1 ⋅ sqrt(x+2) + 3 = 0</span>

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(<span>x+2)</span>)^ 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.
<span>
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
</span>
1⋅ sqrt(x+2)  + 3 = 6

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