Algebra problem involving square roots

[Kevin Licer]

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!

 

[LittleMrsMonkey]

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

 

[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.

 

[Mark44]

So I stumbled upon this problem:
Solve:
View attachment 88442

And I have attempted to solve it, however my solution doesn't match that of the book.
View attachment 88443
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!

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

 

[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.

 

[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?

 

[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.

 

[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.

 

[jk22]

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

 

[Kevin Licer]

Positive?

 

[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 ?

 

[Mark44]

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

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, $\sqrt{x} > 0$.

 

[Kevin Licer]

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

 

[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=?

 

[Kevin Licer]

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