Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

thanks

Answer 5

getting rid of the inequalities made it a lot simpler.

Answer 6

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

Answer 7

analytically, I mean

Answer 8

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

Answer 9

2|-10-1|<10^2 True/False 2|0-1|<0^2 True/False


2|10-1|<10^2 True/False