## sinusoidal 2 years ago 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?

1. 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.

2. myininaya

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

3. sinusoidal

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

4. sinusoidal

thanks

5. sinusoidal

getting rid of the inequalities made it a lot simpler.

6. sinusoidal

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

7. sinusoidal

analytically, I mean

8. 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|<x^2 |dw:1380904061968:dw|

9. 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

10. myininaya

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

11. 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.

12. sinusoidal

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

13. sinusoidal

and I was thrown off by the complex solution too

14. sinusoidal

thanks

15. myininaya

Np. Have fun. :)