# Homework Help: √x-3 + √x = 3 more simple algebra

1. Nov 19, 2007

### viet_jon

[SOLVED] √x-3 + √x = 3 more simple algebra

1. The problem statement, all variables and given/known data

√x-3 + √x = 3

( √x-3 + √x ) ( √x-3 + √x ) = 9

3. The attempt at a solution

I have the answer in my book as x=4 , but I don't understand how they got there.

these are the next two steps which I don't understand, everything after I get so I didn't type it here.

(x-3) + 2√x-3 √x + x = 9

2x - 3 + 2√x²-3x = 9
* the -3x is in the square root

2. Nov 19, 2007

### rock.freak667

(x-3) + 2√x-3 √x + x = 9

you got this by expanding out ( √x-3 + √x ) ( √x-3 + √x )

if you take everything else except the sq.root on the other side(with the 9) you'll get
$\sqrt{x^2-3x}=(6-x)$ then just square both sides and expand again

3. Nov 19, 2007

### viet_jon

man I don't know what you mean. how do you expand out?

4. Nov 19, 2007

### varygoode

Just as in any (a + b)(a + b), you distribute appropriately.

(a + b)(a + b) = (a + b)(a) + (a + b)(b)

= (a)(a) + (b)(a) + (a)(b) + (b)(b).

That's all "expand" means.

5. Nov 19, 2007

### viet_jon

so everytime I have (a + b)(a + b) this is all I have to do is put it in the form you wrote up there? is there anything else I need to know about this?

what is this called anyway? alot of questions i know, sorry....it's just I want to understand it, rather than just memorizing how to do it.

6. Nov 19, 2007

### varygoode

It's just simple distribution. No special name (that I can think of off the top of my head).

As long as you know what it means to distribute, i.e.:

a(b + c) = ab + ac.

7. Nov 19, 2007

### viet_jon

how does (√x-3 + √x) (√x-3 + √x) = 9

distributed into form (a)(a) + (b)(a) + (a)(b) + (b)(b)

= (x-3) + 2√x-3 √x + x = 9 ?

wouldn't it be.... (√x-3)(√x-3) + (√x)(√x-3) + (√x-3)(√x) + (√x)(√x) ?

8. Nov 19, 2007

### varygoode

Your last expression isn't completely simplified.

$$(\sqrt{x-3})(\sqrt{x-3}) = x-3$$

$$(\sqrt{x})(\sqrt{x}) = x$$

And, finally, there are two of the term:

$$(\sqrt{x-3})(\sqrt{x})$$

(which is the same as:)

$$(\sqrt{x})(\sqrt{x-3})$$

which need combining.

That should resolve that particular issue.

9. Nov 20, 2007

### viet_jon

dude...I'm even more lost now. I hate to admit that. I use to be one of the top math students, now I'm struggling with grade 11 crap.

10. Nov 20, 2007

### viet_jon

i hate to keep bothering ya, but I read your last post no less than 20x, and I still don't understand how it works.

11. Nov 20, 2007

### viet_jon

wait...nm.........i got it now.....

thnkx alot brother...owe u 1

12. Nov 20, 2007

### varygoode

Oh, great! Not a problem.

I had started writing a more detailed and fancy response for you, but now I don't need to. Thanks!

Post again if you have any more questions!

P.S.: Instead of triple-posting, try editing your first post next time. :)