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I can't think of any restrictions on squaring both sides of an equation.

Can you square the equation |3|=x
I think if you do that you will loose solutions.

yes, you are right.

But I know that that knowledge is not enough. Can you please provide me some more of such things

I mean a little details, and when not to square etc. etc

I would, but I am very sleepy now. I'll think of some and come back. This is interesting stuff.

Ok thank you

you should ask INewton and Lokisan. I think they will be able to help you the best.

Ok I will do that

Glad that you are here

Could you help please

There are no restrictions.

What is causing you to think that?

You see, I often get more solutions than what I am expected to get, when I square both sides

For example
If I start with sqrt 2 + sqrt 5 =x
sqrt 2 =x - sqrt 5

and then square both sides and go on, I will end up with sqrt 2 +- sqrt 5

So now if we use the formula for quadratic equation

We get two things

right?

Yes

You're getting two answers now because you've asked a different question.

"different question"?

Solving your first equation is not the same as solving the squared one.

I see what you're thinking, but you needn't worry :)

Ok

So does that mean I shouldn't square any equation

I know you're not convinced, but just think about it for a while.

:)

Why did you think you had to square it anyway?

lol

Just working with irrational numbers

You can't get rid of irrational sums by squaring!

Oh...see above.

This made me question, "what are the restrictions..........."

But if you try squaring them, you will find that we get a solution

And you will have to square twice

nope, wolfram alfa says no solutions exist.

If you have time, then I can square both sides, and solve it here

I will do it manually, on paper.

Yes, that is what I am talking of

yes, so solve it here, iamignorant/

sqrt(x+1)-sqrt(x-1)=sqrt(4x-1)

x+1+x-1-2sqrt(x+1)(x-1)=4x-1

2x-2sqrt(x+1)(x-1)=4x-1

-2sqrt(x+1)(x-1)=2x-1

Yes exactly

But how do I know

I mean will I have to plug in every value after solving every equation of the same type

Just what I said.

Does that mean I must always check back?

Yes.

Now what to say......

How about, "I get it"?

sorry?

No...you just learnt something.

I am sorry, but I am still not getting you

Well, it's late, so I'm not really bothered with semantics.

Are you satisfied, Iam?

I will have to be, no other option

I have to go - just think about it...

:Sure I will

Geeeeeez that guy's a buzzkill.

Whatever, he has helped me a lot since the time he joined openstudy

And also is a smart man

That doesn't mean I am trying to offend you Mr Newton

My comment was somewhat tongue-in-cheek.
But, how smart?

Actually I am ignorant, and any comment from my side will simply be a ridiculous one

:)

Any how

Mr Newton, this thing of plugging final solutions, and checking them

What about it?

Should it be done always, or only when we are dealing with square roots

You mean, when I am searching for roots, only then I need to actually do these things

right?

I meant square roots. Roots of the equation would not (normally) have to be checked.

This was the question

If I start with sqrt 2 + sqrt 5 =x
sqrt 2 =x - sqrt 5

I squared both sides and then I solved it

I ended up with sqrt 2 +- sqrt 5

So Mr Lucas commented, that I actually changed the question by squaring it

Could you please provide me a little more explanation

I mean where I changed it, and how did you all guess, that I changed it

Have a nice day.

Yes got you

dis is the definition 4 rational ........i think so

Yes, its okay, it was just a little slip

He meant the other way round

squaring both sides makes the equation quadratic..........which gives an xtra solution......

Which equation are you talking of

u may take any

say x=5

squaring both sides .........
x^2=25
x=5 or -5

bt x=5

ahmm...thnx aditya frm san diego......ryt!!!!!!!!!

I concluded that always I will have to go back to the original equation, and plug in my solutions

jas think b4 squarng

But its hard to accept, although its the truth, as I never did it, all these days.

Very true, but still a bad news for me

Anyhow thank you all

Well, lets come back to the problem I began with

I was trying to prove that \[\sqrt{2}+\sqrt5\]

I am trying to prove that its a irrational number

Lets use the following reasoning
\[[x-\sqrt (2)]^2=5\]

\[x^2-3=2\sqrt2 x\]

\[\sqrt(2)=(x^2-3)(2x)\]

So if I assumed x to be a rational number, then here we get a contradiction

I mean here, its no point plugging back, then what is the fact that allows me to rely on the result

You introduced quite a lot without telling us where it came from (x- sq rt 2)^2=5?

I squared both sides

First I assumed √2+√5=x where x=p/q

Where p and q are co primes

Finally I proved that by that assumption we meet at a contradiction

So I commented that √2+√5 can't be a rational number

Are you getting my question?