What are the restrictions on squaring both sides of an equation.
Please provide a detailed discussion.

- anonymous

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- anonymous

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

- anonymous

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

- anonymous

yes, you are right.

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- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Ok thank you

- anonymous

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

- anonymous

Ok I will do that

- anonymous

Glad that you are here

- anonymous

Could you help please

- anonymous

There are no restrictions.

- anonymous

What is causing you to think that?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I'm not following. If you square both sides, you should get\[(\sqrt{2})^2=(x-\sqrt{5})^2 \rightarrow 2=x^2-2\sqrt{5}x+5\]

- anonymous

So now if we use the formula for quadratic equation

- anonymous

We get two things

- anonymous

right?

- anonymous

Yes

- anonymous

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

- anonymous

"different question"?

- anonymous

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

- anonymous

yes, i think lokisan is right. you asked a different question when you take square root on both sides

- anonymous

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

- anonymous

You have two choices going on here:
1) take your expression as is and solve
2) operate on your expression with an operator that maps this thing to something else and solve for that
When you square, you're doing the latter,

- anonymous

Ok

- anonymous

So does that mean I shouldn't square any equation

- anonymous

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

- anonymous

:)

- anonymous

Why did you think you had to square it anyway?

- anonymous

lol

- anonymous

Just working with irrational numbers

- anonymous

I was trying to prove that the sum of these two irrational numbers will lead to one irrational number

- anonymous

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

- anonymous

Oh...see above.

- anonymous

I have another problem here
sqrt (x+1)-sqrt (x-1)=sqrt (4x-1)
Now if I use wolframalfa, I find no solution
But if I square both sides, I do find

- anonymous

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

- anonymous

Well, I haven't checked it, but if it has no solutions, it just means there are no x-values that will satisfy that expression; it's stuffing up because there are no possible x-values that everything can share that won't make the radicand (thing under sqrt) negative.

- anonymous

Even if you square, you're going to end up with a situation where no x-values are available to make that statement true.
http://www.wolframalpha.com/input/?i=%5Bsqrt+%28x%2B1%29-sqrt+%28x-1%29%5D^2%3Dsqrt+%284x-1%29^2

- anonymous

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

- anonymous

And you will have to square twice

- anonymous

nope, wolfram alfa says no solutions exist.

- anonymous

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

- anonymous

I will do it manually, on paper.

- anonymous

Yes, that is what I am talking of

- anonymous

yes, so solve it here, iamignorant/

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

You do get a solution, x=5/4, but when you test it back in the original equation, the left-hand side is 1, while the right-hand side is 2. This is a contradiction.

- anonymous

Yes exactly

- anonymous

But how do I know

- anonymous

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

- anonymous

You ended up at a contradiction assuming this thing had a solution. The contradiction shows the assumption to be false (reductio ad absurdum).

- anonymous

In general:
You can square an equation and solve as normal, but you must ALWAYS check the solutions in the original, as squaring can produce spurious roots.

- anonymous

Just what I said.

- anonymous

Does that mean I must always check back?

- anonymous

Yes.

- anonymous

Now what to say......

- anonymous

How about, "I get it"?

- anonymous

sorry?

- anonymous

No...you just learnt something.

- anonymous

I am sorry, but I am still not getting you

- anonymous

Sorry I didn't read the whole thing (after you said "There are no restrictions"..., Lokisan).
Regardless, I would not phrase it as you have (with reductio ad absurdum etc...)

- anonymous

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

- anonymous

Are you satisfied, Iam?

- anonymous

I will have to be, no other option

- anonymous

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

- anonymous

:Sure I will

- anonymous

Geeeeeez that guy's a buzzkill.

- anonymous

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

- anonymous

And also is a smart man

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

:)

- anonymous

Any how

- anonymous

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

- anonymous

What about it?

- anonymous

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

- anonymous

There is no need 'normally' if your steps all follow. Only when you have needed to square the equation (i.e. basically when you have roots etc)

- anonymous

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

- anonymous

right?

- anonymous

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

- anonymous

If you notice some of our conversation at the beginning, you will notice, that once Mr Lucas said that I have actually changed my question by squaring both sides

- anonymous

This was the question

- anonymous

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

- anonymous

I squared both sides and then I solved it

- anonymous

I ended up with sqrt 2 +- sqrt 5

- anonymous

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

- anonymous

Could you please provide me a little more explanation

- anonymous

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

- anonymous

OK. Consider the follow equations:
\[\sqrt{3x^2+1} + \sqrt{x} - 2x -1 = 0 \]
\[\sqrt{3x^2+1} -2\sqrt{x} +x - 1 = 0\]
\[\sqrt{3x^2+1} -2\sqrt{x} -x + 1 = 0\]
To solve these, you will need to square them (once if you split it up smartly onto each side it smart, but maybe twice if you don't).
This will give the same 3 roots for all equations (which it is left for you to find if you are bored).
However, they each have different solutions. Why?
BECAUSE SQUARING CAN (but will not always) PRODUCE SPURIOUS ROOTS, which must be checked and rejected. The algebraic manipulation (in this case, squaring), has changed what you are trying to solve, and given extra roots.

- anonymous

If you only read one part, read the last paragraph. It will not ALWAYS change the values that solve it, but it can so they MUST be checked.

- anonymous

Have a nice day.

- anonymous

Yes got you

- anonymous

After reading above, I see you were trying to prove irrational + irrational = irrational.
This is not (always) true. Counterexample:
\[\pi + (-\pi) = 0 \not= \text{irrational} \]
But it can be. In general just use the definition of irrational (i.e can be written p/q with p and q coprime integers) and use that fact.

- anonymous

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

- anonymous

Yes, its okay, it was just a little slip

- anonymous

He meant the other way round

- anonymous

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

- anonymous

Which equation are you talking of

- anonymous

u may take any

- anonymous

say x=5

- anonymous

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

- anonymous

bt x=5

- anonymous

yes, siddharth is right. Squaring both sides makes it a quadratic equation, providing an extra or as Inewton called it, a spurious root.

- anonymous

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

- anonymous

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

- anonymous

jas think b4 squarng

- anonymous

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

- anonymous

1) bah, yes, slight error on the definition of irrational, but you guys sorted it :)
2) It's not as simple as 'think before squaring' - your example was nice as it gave an obvious example of when an extra root is produced. It will almost always never be that nice. If you see my examples, it is not 'obvious' which roots you produce are not valid etc. So yes, just check them.

- anonymous

Very true, but still a bad news for me

- anonymous

Anyhow thank you all

- anonymous

Sometimes when squaring you can introduce extraneous solution. I always remember asking my trig teacher how come such and such was not result; he explained how an 'extraneous' solution was introduced when I squared.

- anonymous

Well, lets come back to the problem I began with

- anonymous

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

- anonymous

I am trying to prove that its a irrational number

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

But now my question is, as we came to this conclusion by squaring both sides, how do I be sure, that the equation is speaking the truth.

- anonymous

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

- anonymous

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

- anonymous

I squared both sides

- anonymous

First I assumed √2+√5=x where x=p/q

- anonymous

Where p and q are co primes

- anonymous

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

- anonymous

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

- anonymous

Now during this operation I squared both sides, and as squaring both sides leads to a non reliable equation, then how can I rely on the result

- anonymous

Are you getting my question?

- anonymous

Your reasoning looks OK (yours steps just follow after one another - if this is rational, then this squared is rational... etc.; it is not the same problem as when trying to solve equations after squaring)
On a side note, I would personally argue it thus (where p and q are blah blah..):
\[\text{Assume } \sqrt{5} + \sqrt{2} = \frac{p}{q} = \text{rational}\]
\[\implies 7+2\sqrt{10} = \left(\frac{p}{q}\right)^2 = \text{rational}\]
\[\implies 2\sqrt{10} = \left(\frac{p}{q}\right)^2 -7 = \text{rational} \]
And showing sqrt(10) is irrational is just old hat (or maybe can be assumed, as yours does with sqrt(2) ). Therefore out assumption was false etc.
But it's pretty much the same thing :(

- anonymous

May be this time I am sounding idiotic, but the sentence "it is not the same problem as when trying to solve equations after squaring" to be frank has not satisfied me

- anonymous

I really don't understand why I should rely on an equation which is the result of squaring both sides, when I have seen earlier that it results in spurious results

- anonymous

I don't know about the proof, but on the way you made a mistake in algebra from this line x^2 - 3=2x sq rt 2 does not equal sq rt 2=(x^2-3)2x

- anonymous

You are not trying to solve for x. If (z) is rational, than z^2 is rational. That is the only assumption you are making. It is a completely different circumstance (at least when used in my proof; yours splits it up slightly so may not be viewed in the same way).

- anonymous

So I conclude that since I am not interested in the number of solutions, and since the assumption doesn't get hampered by squaring, I can go on. Right?

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