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- anonymous

2|x-1| < x^2
answers are: x < -1 - sqrt(3) or x > -1+sqrt(3)
I'm having trouble coming up with these solutions. Could someone help me with the steps?

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- anonymous

- katieb

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- myininaya

Did you try looking at 2|x-1|=x^2 first
When we have |f(x)|=a, we try to solve this by doing f(x)=-a of f(x)=a
Now keep in mind that a needs to be positive or 0.
Guess what? x^2 is a positive number or 0.

- myininaya

I'm tell you to solve the following
2(x-1)=x^2 or 2(x-1)=-x^2

- anonymous

I've got it now. I thought that I tried that before, but I think I was just making so many mistakes that I became too sloppy and frustrated

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- anonymous

thanks

- anonymous

getting rid of the inequalities made it a lot simpler.

- anonymous

how would you know how to replace the inequalities back into the answer?

- anonymous

analytically, I mean

- myininaya

One of the equations you solve, you will get a complex answer.
The other equation you solve, will give you two real solutions. You should see x=-1+sqrt(3) or x=-1-sqrt(3).
You can test intervals to see where we have 2|x-1|

- myininaya

|dw:1380904086865:dw|
2|-10-1|<10^2 True/False 2|0-1|<0^2 True/False
2|10-1|<10^2 True/False

- myininaya

The inequalities that come back true are the intervals you want to include in your solution.

- myininaya

The first inequality, I went ahead and said 10^2 instead of (-10)^2 since 10^2 is the same result at (-10)^2.

- anonymous

right, I get it now. I wasn't thinking about getting rid of the inequalities and just replacing them back at the end. it makes the process a lot simpler

- anonymous

and I was thrown off by the complex solution too

- anonymous

thanks

- myininaya

Np. Have fun. :)

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