Mathematics

6:02 PM

@MichaelGreinecker,Thanks! I will keep that in mind. It seems my basic stuff can be learnt from Abbott's Understanding Analysis

Michael Greinecker

@Sabyasachi That is certainly a nice book to start with and very readable.

Sabyasachi

looks like I have to run. bye.

Thanks again

Michael Greinecker

bye

robjohn

@JohnJunior: You afraid to leave my smiley? >8(

John Junior

No, Sir.
$\Huge\text{>8(}$

Mariano Suárez-Alvarez

6:12 PM

«Vowels are people too» is the funniest thing I've read on MSE ever

John Senior

I once read a suggestion that since the Czech language seems to be short of vowels and Irish seems to have too many, there might be a business opportunity - smuggling vowels from Ireland to the Czech Republic :)

John Junior

What will they think of next?

>8(

Matt

6:31 PM

@robjohn Thanks, it's been done already.

robjohn

@Matt Okay. Sorry to take so long.

Dan the Man

6:47 PM

Hello

Can $2y\sqrt{20}{5}$ be reduced any further?

arrrg That's supposed to be the fifth root of 20

Matt

Is this a duplicate?

@robjohn Can we repeat the game though? I don't see why this has 3 votes as "off topic". Seems very on topic to me.

Dan the Man

7:03 PM

How do you solve this? $\sqrt{n} + 1 = 4$

Jeff

@DantheMan first, subtract 1 from both sides. What do you get then?

Dan the Man

$sqrt{n} = 3$

* $\sqrt{n} = 3$

Jeff

@DantheMan Good, now you can square both sides of the equation (right?). What do you get then?

Dan the Man

You can square both sides?

Jeff

@Dan Yes!

Dan the Man

7:05 PM

Ahhhh

n = 9

Jeff

but you must square the whole side! and you must do the same thing to both sides (you knew that one already, probably).

@DantheMan And that's the answer. Plug it back into the original equation to see if it works out right.

Dan the Man

Awesome! Thanks

Jeff

@Dan YW. glad I could help

Dan the Man

Just to make sure... In $\sqrt{x + d^2} = c$

...Solving for $x$ gives $x = c^2$ ?

Wait... that's wrong

$x = c - d^2$ ?

anon

$\sqrt{x+d^2}=c\implies x+d^2=c^2\implies x=c^2-d^2$.

Dan the Man

7:14 PM

Right... Duh! >.<

I hate making simple mistakes in Algebra

Here's another

$\sqrt{m+4} = 4 - \sqrt{m-4}$

Jeff

@DantheMan that's a little more complicated now. If you try to square both sides of the equation, you will get a more complicated problem (have you tried that?).

Dan the Man

Yes... It didn't work.

Jeff

squaring both sides for that equation is also very error prone. Do you see a way to simplify that equation before squaring both sides?

Dan the Man

Isolating the variable?

Jeff

@Dan oh, wait. Is one of the radicals $m+4$ and the other one $m-4$?

Dan the Man

7:21 PM

Precisely

Jeff

@DantheMan good thought, but you won't be able to isolate the variable since it is under the radical sign.

ok. hold on a sec then (although anon might do this equation quickly for you, but i have to think about it a sec).

Dan the Man

Hah alright

Can't I just add $\sqrt{m-4}$ to both sides?

Jeff

@Dan that's what i'm thinking of trying. And then one side is $\sqrt{m+4}+\sqrt{m-4}$. Then square that side. But I want to see what happens before I told you to do that.

Dan the Man

Hmmmm

Ah, when I square both sides...

I will get

$(\sqrt{m+4}) = (4-\sqrt{m-4})^2$

So the left side will have to be FOILed, right?

*right side, that is

Alright let me try

Jeff

@Dan I retract that statement. You're way worked out easier!

Dan the Man

7:30 PM

$m=2$ ?

Jeff

@DantheMan yes, RHS will have to be foiled. Also, don't forget to square the LHS (you forgot to type the exponent).

Dan the Man

Ah cool!

Jeff

plug $m=2$ back in. I think you made a mistake somewhere.

Dan the Man

Hmmm.. I got $m=2$ with your method

What did you get using my method?

Jeff

I'm not tellin' (yet). :D Show me the first step using my method.

Dan the Man

7:33 PM

haha ok

Jeff

you shoudl have squared $\sqrt{m+4}+\sqrt{m-4}$, right?

Dan the Man

Ahhh I went off there

Jeff

@Dan is this homework? I will make you do at least a little work before giving a straight out answer.

Dan the Man

I just took off the radicals Lol

Lol yeah it's homework

Jeff

@DantheMan <smh> lol

Dan the Man

7:35 PM

<smh> ?

Jeff

smh=shakes my head... it was done in good fun (with a smile) since you had already "lol"ed

:D

Dan the Man

Hahaha

Jeff

try again... but use your method. i'm trying my method now and i just keep getting deeper and deeper into non-simplifying equations)

Dan the Man

Hmmm ok

John Senior

Can I add my ideas?

Dan the Man

7:38 PM

I can't even remember my method... LOL

@JohnSenior Yes please

John Senior

We are trying to solve $\sqrt{m+4} = 4 - \sqrt{m-4}$ ?

Dan the Man

Indeed

Jeff

@JohnSenior of course.

John Senior

OK - squaring both sides gives $(\sqrt{m+4})^2 = (4 - \sqrt{m-4})^2$

Jeff

@Dan your idea was to square both sides right away. (mine was to move the radicals to the same side - mine made the problem more complicated).

Dan the Man

7:39 PM

Ah right.

John Senior

which simplifies to $m+4 = 16 - 8\sqrt{m-4} + (m-4)$

Dan the Man

Right

John Senior

now move the square root to one side and everything else to the other side

and get $8\sqrt{m-4} = 8$

or $\sqrt{m-4} = 1$

Dan the Man

Yes!

John Senior

so $m-4 = 1$, and finally $m=5$ - then check that this works by substituting back into the original equation

Jeff

7:46 PM

@John that's where I was going! :D (i just wanted to make him do most of the work, which takes more time) :D

Dan the Man

Awesome

@Jeff Hey!

Jeff

as an aside: does the square root sign look funny today?

John Senior

@DantheMan - now you need to do a few more of the same type to make sure you can do the whole process yourself

Dan the Man

@JohnSenior Yeah

@Jeff ???

John Senior

you will find there are a few quirks in some examples that look similar, due to the fact that squaring both sides of an equation might introduce extra solutions which do not satisfy the original equation

Jeff

7:48 PM

is the top of the square root symbol sagging on your monitor?

Dan the Man

@Jeff No?

John Senior

Looks fine here - but some people have reported problems with some browsers/OSes

Dan the Man

@JohnSenior Ah ok

Jeff

I'm using Chrome/Win-doze.

John Senior

@DantheMan here are some examples to practice on

Dan the Man

7:57 PM

@JohnSenior Thanks

John Senior

@DantheMan Have fun - and remember always to check your solutions

Ramana Venkata

Does anybody know about percolation theory here??

Dan the Man

@JohnSenior Yeah

Ed Gorcenski

8:18 PM

@Jeff In chrome, I get many issues with sqrt bars.

Jeff

that's too bad. cuz i like chrome so much better than the other browsers

Ed Gorcenski

Me too; well, I used to. I have found it to be very slow lately when going through my firewall at the office. I worried about the information it was sending out.

I've yet to find a decent replacement, though.

Jeff

@Ed re: your last statement. that's the bottom line. there are no good browsers (just like politicians). you use the least bad.

Dan the Man

How do I find this?

Which is greater, $\sqrt{3}$ or $\sqrt{3}{5}$ ?

Wait...

How do I do cubed roots in MathJax?

Transcript for

$Mathematics$ Mathematics