Quick question, say I'm simplying a solution I got using the quadratic equation and I run into this:

Original version (as posted by OP):

x = -7 +- 3 sqrt(5) over 3

Edited version: $$x = \frac{-7\pm 3 \sqrt{5} }{3}$$ Would the two $3$s cross each out leaving the answer to be $x = -7 \pm \sqrt{5}$, or is that illegal in terms of rules and you have to simplify all the terms, including the $-7$ if you were to simplify correctly.

Thanks!

-
$\frac{a\pm b}{3} = \frac a 3 \pm \frac b 3$ – lhf Dec 6 '11 at 11:58
Does my eq. edit correspond to what you wanted to ask? – user13838 Dec 6 '11 at 12:20
@percusse: your edit makes Gerry's first sentence (and indeed the whole answer) look rather silly. Please do not edit in such ambiguous cases but rather suggest your interpretation in a comment. – t.b. Dec 6 '11 at 12:23
@t.b. It says quadratic equation so it doesn't take too much to guess what OP means, that's why I asked him (it doesn't take too much effort to rollback anyway). And I don't think any edit can make Gerry's arguments silly. – user13838 Dec 6 '11 at 12:27
@percusse: Unfortunately you've removed the opportunity for us to impress on OP the importance of parentheses... – J. M. Dec 6 '11 at 12:29

I'm a little late on the scene, but from your most recent comments you still haven't seen the light. Maybe this will help.

$$\frac{-7\pm 3 \sqrt{5} }{3} \;\; =\;\; \left(\frac{1}{3}\right) \left(\frac{-7\pm 3 \sqrt{5}}{1}\right) \;\;= \;\; \left(\frac{1}{3}\right)\left(-7 \; \pm \; 3 \sqrt{5}\right)$$

$$= \;\; \left(\frac{1}{3}\right)(-7) \; \pm \; \left(\frac{1}{3}\right)\left(3 \sqrt{5}\right) \;\; = \;\; \left(\frac{1}{3}\right)\left(\frac{-7}{1}\right) \; \pm \; \left(\frac{1}{3}\right)\left(\frac{3 \sqrt{5}}{1}\right) \;\; = \;\; -\frac{7}{3}\; \pm \; \frac{3 \sqrt{5}}{3}$$

Usually people do all this in one step (see the first comment under your question, the comment by lhf) and write:

$$\frac{-7\pm 3 \sqrt{5} }{3} \;\; = \;\; -\frac{7}{3}\; \pm \; \frac{3 \sqrt{5}}{3}$$

-
People usually do a little bit more, simplifying to $$-{7\over3}\pm\sqrt5$$ but I think what you've done should be most helpful to OP. – Gerry Myerson Dec 7 '11 at 3:44

What you have written is ambiguous. You might mean $x=-7+(3\sqrt5/3)$, but I bet you mean $x=(-7+3\sqrt5)/3$. So let me ask you: in $(1+2)/2$, can you "cancel the 2s" to get $(1+2)/2=(1+1)/1=2/1=2$?

-
I don't think so, would you be able to if the (1 + 2) weren't in parathesis? – user1062058 Dec 6 '11 at 12:10
The trouble with leaving out the parentheses is that only you would know whether $1+2/2$ means $(1+2)/2$ or $1+(2/2)$. Everyone else would have to guess which one you meant. That's why we use parentheses; we don't want to make people guess what we mean. With $1+(2/2)$, sure, you can cancel. – Gerry Myerson Dec 6 '11 at 12:15
Okay, with that in mind. Is the answer to my question indeed x=−7±√5 because I can cancel the 3s but leave the 7 as is? – user1062058 Dec 6 '11 at 12:25
Look: what you wrote in your question is ambiguous. No one can tell you the answer, because only you know what you mean by what you wrote, because you haven't put in any parentheses. If you put in parentheses, people can answer your question, but you are the only one who knows where the parentheses belong, so you have to put them in. – Gerry Myerson Dec 6 '11 at 12:30
Someone edited my question, there is no parentheses in it at all, can you look at the question again? Thanks for all the help so far – user1062058 Dec 6 '11 at 12:36