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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can't think of any restrictions on squaring both sides of an equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you square the equation 3=x I think if you do that you will loose solutions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But I know that that knowledge is not enough. Can you please provide me some more of such things

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean a little details, and when not to square etc. etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would, but I am very sleepy now. I'll think of some and come back. This is interesting stuff.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you should ask INewton and Lokisan. I think they will be able to help you the best.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Glad that you are here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could you help please

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are no restrictions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is causing you to think that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You see, I often get more solutions than what I am expected to get, when I square both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For example If I start with sqrt 2 + sqrt 5 =x sqrt 2 =x  sqrt 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then square both sides and go on, I will end up with sqrt 2 + sqrt 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not following. If you square both sides, you should get\[(\sqrt{2})^2=(x\sqrt{5})^2 \rightarrow 2=x^22\sqrt{5}x+5\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So now if we use the formula for quadratic equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're getting two answers now because you've asked a different question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"different question"?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solving your first equation is not the same as solving the squared one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, i think lokisan is right. you asked a different question when you take square root on both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see what you're thinking, but you needn't worry :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0So does that mean I shouldn't square any equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know you're not convinced, but just think about it for a while.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why did you think you had to square it anyway?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just working with irrational numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was trying to prove that the sum of these two irrational numbers will lead to one irrational number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can't get rid of irrational sums by squaring!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have another problem here sqrt (x+1)sqrt (x1)=sqrt (4x1) Now if I use wolframalfa, I find no solution But if I square both sides, I do find

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This made me question, "what are the restrictions..........."

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I haven't checked it, but if it has no solutions, it just means there are no xvalues that will satisfy that expression; it's stuffing up because there are no possible xvalues that everything can share that won't make the radicand (thing under sqrt) negative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Even if you square, you're going to end up with a situation where no xvalues are available to make that statement true. http://www.wolframalpha.com/input/?i=%5Bsqrt+%28x%2B1%29sqrt+%28x1%29%5D^2%3Dsqrt+%284x1%29^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But if you try squaring them, you will find that we get a solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And you will have to square twice

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope, wolfram alfa says no solutions exist.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you have time, then I can square both sides, and solve it here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I will do it manually, on paper.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, that is what I am talking of

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, so solve it here, iamignorant/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(x+1)sqrt(x1)=sqrt(4x1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x+1+x12sqrt(x+1)(x1)=4x1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02x2sqrt(x+1)(x1)=4x1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02sqrt(x+1)(x1)=2x1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You do get a solution, x=5/4, but when you test it back in the original equation, the lefthand side is 1, while the righthand side is 2. This is a contradiction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean will I have to plug in every value after solving every equation of the same type

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You ended up at a contradiction assuming this thing had a solution. The contradiction shows the assumption to be false (reductio ad absurdum).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In 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
 5 years ago
Best ResponseYou've already chosen the best response.0Does that mean I must always check back?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now what to say......

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How about, "I get it"?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No...you just learnt something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am sorry, but I am still not getting you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry 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
 5 years ago
Best ResponseYou've already chosen the best response.0Well, it's late, so I'm not really bothered with semantics.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you satisfied, Iam?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I will have to be, no other option

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have to go  just think about it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Geeeeeez that guy's a buzzkill.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whatever, he has helped me a lot since the time he joined openstudy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And also is a smart man

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That doesn't mean I am trying to offend you Mr Newton

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My comment was somewhat tongueincheek. But, how smart?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually I am ignorant, and any comment from my side will simply be a ridiculous one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mr Newton, this thing of plugging final solutions, and checking them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Should it be done always, or only when we are dealing with square roots

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There 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
 5 years ago
Best ResponseYou've already chosen the best response.0You mean, when I am searching for roots, only then I need to actually do these things

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I meant square roots. Roots of the equation would not (normally) have to be checked.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0This was the question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If I start with sqrt 2 + sqrt 5 =x sqrt 2 =x  sqrt 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I squared both sides and then I solved it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I ended up with sqrt 2 + sqrt 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So Mr Lucas commented, that I actually changed the question by squaring it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could you please provide me a little more explanation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean where I changed it, and how did you all guess, that I changed it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK. 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
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0After 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
 5 years ago
Best ResponseYou've already chosen the best response.0dis is the definition 4 rational ........i think so

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, its okay, it was just a little slip

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0He meant the other way round

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0squaring both sides makes the equation quadratic..........which gives an xtra solution......

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which equation are you talking of

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0squaring both sides ......... x^2=25 x=5 or 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, siddharth is right. Squaring both sides makes it a quadratic equation, providing an extra or as Inewton called it, a spurious root.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahmm...thnx aditya frm san diego......ryt!!!!!!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I concluded that always I will have to go back to the original equation, and plug in my solutions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But its hard to accept, although its the truth, as I never did it, all these days.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01) 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
 5 years ago
Best ResponseYou've already chosen the best response.0Very true, but still a bad news for me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sometimes 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
 5 years ago
Best ResponseYou've already chosen the best response.0Well, lets come back to the problem I began with

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was trying to prove that \[\sqrt{2}+\sqrt5\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am trying to prove that its a irrational number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lets use the following reasoning \[[x\sqrt (2)]^2=5\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt(2)=(x^23)(2x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So if I assumed x to be a rational number, then here we get a contradiction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But 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
 5 years ago
Best ResponseYou've already chosen the best response.0I mean here, its no point plugging back, then what is the fact that allows me to rely on the result

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You introduced quite a lot without telling us where it came from (x sq rt 2)^2=5?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First I assumed √2+√5=x where x=p/q

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where p and q are co primes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Finally I proved that by that assumption we meet at a contradiction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I commented that √2+√5 can't be a rational number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now 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
 5 years ago
Best ResponseYou've already chosen the best response.0Are you getting my question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your 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
 5 years ago
Best ResponseYou've already chosen the best response.0May 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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^23)2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0So 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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