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- anonymous

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- chestercat

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- anonymous

Task 2
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
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.

- anonymous

An extraneous solution is a fake result. You can go through and solve the equation, but when you check the answer it won't work.
The easiest thing to do for both equations is to make a = 1 so you don't have to worry about division. Then you're left with \(\sqrt{x+b}+c=d\)
Subtract c from both sides to get \(\sqrt{x+b}=d-c\)
For the extraneous solution equation, you need d-c to be negative. For the other one d-c will be positive.

- anonymous

so I can choose any number or does It have to be specific

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## More answers

- anonymous

You can choose any number

- anonymous

ok

- anonymous

so I plug in the numbers ad solve

- anonymous

I couldn't see anything in that file. Can you post a screenshot or use the draw tool?

- anonymous

ok I plugged In numbers a radical sign 18 +24+32=46

- anonymous

so
I subtract the 32 from both sides now right

- anonymous

you're still supposed to have a radical sign and an x. The only numbers you're replacing are a,b,c, and d. For example
√(x-3)+4=20

- anonymous

For that equation a =1, b=-3, c = 4, and d = 20.
Check d - c:
d - c = 20 - 4 = 16
This is positive so that's an equation without an extraneous solution

- anonymous

So I don't need to add a number for x

- anonymous

right. x stays

- anonymous

So so I subtract 4 from both sides and then now I have 1\ x+3=16 sorry I don't have a radical sign on my keyboard lol

- anonymous

So I keep simplifying the equation from there

- anonymous

yes after you subtract 4 square both sides

- anonymous

Ok now I have 1/ x + 3 =16

- anonymous

Than I subtracted 3 from both sides and now it's just 1/x=13

- anonymous

no, square both sides first then subtract 3. You can't subtract the 3 while it's under the radical.|dw:1437332991847:dw|

- anonymous

Square the 3 also or no

- anonymous

no, the whole thing under the radical is squared. The √ and ² cancel each other
\[\sqrt{x-3}^2=x-3\]

- anonymous

Oh ok so x + 3 =256 so do I subtract the three now or leave it as it is

- anonymous

subtract the 3

- anonymous

x=233

- anonymous

253?
Just realized we switched for x-3 to x+3 when we were solving it. Just change the sign in the original equation to +

- anonymous

Ok lol yea it's 253

- anonymous

For the extraneous equation, just swap c and d and go through the same steps
√(x-3)+20=4

- anonymous

Ok I just finished that part now how would I explain why the first one was extraneous do I just say because the answer doesn't go with the equation

- anonymous

yes, the answer doesn't check

- anonymous

Ok thank you

- anonymous

you're welcome

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