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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First question answer: The limit is 69

Second question answer: The limit is 5

Step-by-step explanation:

For the first limit, plug in $x=8$ in the expression $(9x-3)$ , that's the answer for linear equations and limits.

So we have:

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The answer is 69

For the second limit, if we do same thing as the first, we will get division by 0. Also indeterminate form, 0 divided by 0. Thus we would think that the limit does not exist. But if we do some algebra, we can easily simplify it and thus plug in the value $x=1$ into the simplified expression to get the correct answer. Shown below: