anonymous
  • 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?
Mathematics
katieb
  • katieb
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myininaya
  • 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
  • myininaya
I'm tell you to solve the following 2(x-1)=x^2 or 2(x-1)=-x^2
anonymous
  • 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
  • anonymous
thanks
anonymous
  • anonymous
getting rid of the inequalities made it a lot simpler.
anonymous
  • anonymous
how would you know how to replace the inequalities back into the answer?
anonymous
  • anonymous
analytically, I mean
myininaya
  • 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
  • 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
  • myininaya
The inequalities that come back true are the intervals you want to include in your solution.
myininaya
  • 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
  • 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
  • anonymous
and I was thrown off by the complex solution too
anonymous
  • anonymous
thanks
myininaya
  • myininaya
Np. Have fun. :)

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