# Some algebra

• June 24th 2010, 10:18 AM
maggiec
Some algebra
Here is the question:

Attachment 17986

I've done part a), however in part b) I'm confused as how to get rid of the modulus..

Any help is appreciated!

Thanks
• June 24th 2010, 10:37 AM
Unknown008

After making part a), you'll see that there are two points of intersection.

First, solve for x being positive, rejecting one value. Your diagram would tell you which one to reject.
Then, do the same thing with the negative value of x.

Finally, you should get

x = 2

$x = \frac14(-1-\sqrt{17})$
• June 24th 2010, 11:07 AM
maggiec
Quote:

Originally Posted by Unknown008

After making part a), you'll see that there are two points of intersection.

First, solve for x being positive, rejecting one value. Your diagram would tell you which one to reject.
Then, do the same thing with the negative value of x.

Finally, you should get

x = 2

$x = \frac14(-1-\sqrt{17})$

I got both results thanks!
• June 24th 2010, 11:11 AM
Unknown008
You're welcome(Happy)