# Homework Help: Algebra problem involving square roots

1. Sep 9, 2015

### Kevin Licer

• Member warned about posting with no template
So I stumbled upon this problem:
Solve:

And I have attempted to solve it, however my solution doesn't match that of the book.

The solution should be x+4, but I don't have a clue how to get to that. I'm sorry if this seems pretty rudimentary, but I just want some help because it's the first time I've dealt with an equation like this (1<x<3 what should I do with that?). Any help is appreciated. Thanks!

2. Sep 9, 2015

### LittleMrsMonkey

When the last sq.root goes,what remains is the absolute value,so you are left woth (-x+3) because x<3

3. Sep 9, 2015

### Kevin Licer

So the solution is going to be |3x-2|? Sorry if I'm not getting this, but could you please do it step by step, I'm really bad at this.

4. Sep 9, 2015

### Staff: Mentor

Technically, you aren't "solving" anything -- that term is applied to equations or inequalities. What you are doing is simplifying the given expression.
$$\sqrt{(x + 2)^2} + \sqrt{(x - 1)^2} + \sqrt{(x - 3)^2}, 1 < x < 3$$
$$= |x + 2| + |x - 1| + |x - 3|$$
You need to take the absolute value expressions on the given interval, 1 < x < 3, into consideration.

5. Sep 9, 2015

### jk22

What is meant is that $$\sqrt{x^2}=|x|$$

Then you have to check if your x in your case x+2, x-1, x-3 are positve or negative in the domain 1<x<3.

Last edited: Sep 9, 2015
6. Sep 9, 2015

### Kevin Licer

So, because we need to take the given interval into consideration, we take the absolute value of each term? So that would be the final "simplification" and not x+4?

7. Sep 9, 2015

### jk22

If you consider only the interval given it will give x+4

But on other intervals the solution will be different.

And sometimes we have to choose the sign of the square root depending on the definition.

8. Sep 9, 2015

### Kevin Licer

But how do I check if my x in x+2, x-1, x-3 is positive or negative? And could you give an example of how it would be if the domain was slightly different? Apologies if I may be annoying you with too many questions, but I want to truly understand.

9. Sep 9, 2015

### jk22

For example if 1<x<3 is x+2 positive or negative ?

10. Sep 9, 2015

### Kevin Licer

Positive?

11. Sep 9, 2015

### jk22

Yes it runs from 3 to 5, so |x+2|=x+2 for x in that domain.

Now you do the same reasoning for x-1 and x-3 so you find ?

12. Sep 9, 2015

### Staff: Mentor

I'm not sure what you're saying here. By definition and long usage, the square root of a positive number is positive. In symbols, if x > 0, $\sqrt{x} > 0$.

13. Sep 9, 2015

### Kevin Licer

So what happens if it's negative, for x-1 and x-3? How would I do it now?

14. Sep 9, 2015

### jk22

You notice that x-1 is positive whereas x-3 is negative for 1<x<3

Hence |x-1|=x-1 and |x-3|=-(x-3)=3-x

Then you make the sum : x+2+x-1+3-x=?

15. Sep 9, 2015

### Kevin Licer

Oh, now I see. Well thank you and everyone else for clarifying and helping me understand this problem. Thanks a bunch!