anonymous
  • anonymous
What are the restrictions on squaring both sides of an equation. Please provide a detailed discussion.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I can't think of any restrictions on squaring both sides of an equation.
anonymous
  • anonymous
Can you square the equation |3|=x I think if you do that you will loose solutions.
anonymous
  • anonymous
yes, you are right.

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anonymous
  • anonymous
But I know that that knowledge is not enough. Can you please provide me some more of such things
anonymous
  • anonymous
I mean a little details, and when not to square etc. etc
anonymous
  • anonymous
I would, but I am very sleepy now. I'll think of some and come back. This is interesting stuff.
anonymous
  • anonymous
Ok thank you
anonymous
  • anonymous
you should ask INewton and Lokisan. I think they will be able to help you the best.
anonymous
  • anonymous
Ok I will do that
anonymous
  • anonymous
Glad that you are here
anonymous
  • anonymous
Could you help please
anonymous
  • anonymous
There are no restrictions.
anonymous
  • anonymous
What is causing you to think that?
anonymous
  • anonymous
You see, I often get more solutions than what I am expected to get, when I square both sides
anonymous
  • anonymous
For example If I start with sqrt 2 + sqrt 5 =x sqrt 2 =x - sqrt 5
anonymous
  • anonymous
and then square both sides and go on, I will end up with sqrt 2 +- sqrt 5
anonymous
  • 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
  • anonymous
So now if we use the formula for quadratic equation
anonymous
  • anonymous
We get two things
anonymous
  • anonymous
right?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
You're getting two answers now because you've asked a different question.
anonymous
  • anonymous
"different question"?
anonymous
  • anonymous
Solving your first equation is not the same as solving the squared one.
anonymous
  • anonymous
yes, i think lokisan is right. you asked a different question when you take square root on both sides
anonymous
  • anonymous
I see what you're thinking, but you needn't worry :)
anonymous
  • 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
  • anonymous
Ok
anonymous
  • anonymous
So does that mean I shouldn't square any equation
anonymous
  • anonymous
I know you're not convinced, but just think about it for a while.
anonymous
  • anonymous
:)
anonymous
  • anonymous
Why did you think you had to square it anyway?
anonymous
  • anonymous
lol
anonymous
  • anonymous
Just working with irrational numbers
anonymous
  • anonymous
I was trying to prove that the sum of these two irrational numbers will lead to one irrational number
anonymous
  • anonymous
You can't get rid of irrational sums by squaring!
anonymous
  • anonymous
Oh...see above.
anonymous
  • 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
  • anonymous
This made me question, "what are the restrictions..........."
anonymous
  • 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
  • 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
  • anonymous
But if you try squaring them, you will find that we get a solution
anonymous
  • anonymous
And you will have to square twice
anonymous
  • anonymous
nope, wolfram alfa says no solutions exist.
anonymous
  • anonymous
If you have time, then I can square both sides, and solve it here
anonymous
  • anonymous
I will do it manually, on paper.
anonymous
  • anonymous
Yes, that is what I am talking of
anonymous
  • anonymous
yes, so solve it here, iamignorant/
anonymous
  • anonymous
sqrt(x+1)-sqrt(x-1)=sqrt(4x-1)
anonymous
  • anonymous
x+1+x-1-2sqrt(x+1)(x-1)=4x-1
anonymous
  • anonymous
2x-2sqrt(x+1)(x-1)=4x-1
anonymous
  • anonymous
-2sqrt(x+1)(x-1)=2x-1
anonymous
  • 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
  • anonymous
Yes exactly
anonymous
  • anonymous
But how do I know
anonymous
  • anonymous
I mean will I have to plug in every value after solving every equation of the same type
anonymous
  • 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
  • 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
  • anonymous
Just what I said.
anonymous
  • anonymous
Does that mean I must always check back?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Now what to say......
anonymous
  • anonymous
How about, "I get it"?
anonymous
  • anonymous
sorry?
anonymous
  • anonymous
No...you just learnt something.
anonymous
  • anonymous
I am sorry, but I am still not getting you
anonymous
  • 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
  • anonymous
Well, it's late, so I'm not really bothered with semantics.
anonymous
  • anonymous
Are you satisfied, Iam?
anonymous
  • anonymous
I will have to be, no other option
anonymous
  • anonymous
I have to go - just think about it...
anonymous
  • anonymous
:Sure I will
anonymous
  • anonymous
Geeeeeez that guy's a buzzkill.
anonymous
  • anonymous
Whatever, he has helped me a lot since the time he joined openstudy
anonymous
  • anonymous
And also is a smart man
anonymous
  • anonymous
That doesn't mean I am trying to offend you Mr Newton
anonymous
  • anonymous
My comment was somewhat tongue-in-cheek. But, how smart?
anonymous
  • anonymous
Actually I am ignorant, and any comment from my side will simply be a ridiculous one
anonymous
  • anonymous
:)
anonymous
  • anonymous
Any how
anonymous
  • anonymous
Mr Newton, this thing of plugging final solutions, and checking them
anonymous
  • anonymous
What about it?
anonymous
  • anonymous
Should it be done always, or only when we are dealing with square roots
anonymous
  • 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
  • anonymous
You mean, when I am searching for roots, only then I need to actually do these things
anonymous
  • anonymous
right?
anonymous
  • anonymous
I meant square roots. Roots of the equation would not (normally) have to be checked.
anonymous
  • 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
  • anonymous
This was the question
anonymous
  • anonymous
If I start with sqrt 2 + sqrt 5 =x sqrt 2 =x - sqrt 5
anonymous
  • anonymous
I squared both sides and then I solved it
anonymous
  • anonymous
I ended up with sqrt 2 +- sqrt 5
anonymous
  • anonymous
So Mr Lucas commented, that I actually changed the question by squaring it
anonymous
  • anonymous
Could you please provide me a little more explanation
anonymous
  • anonymous
I mean where I changed it, and how did you all guess, that I changed it
anonymous
  • 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
  • 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
  • anonymous
Have a nice day.
anonymous
  • anonymous
Yes got you
anonymous
  • 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
  • anonymous
dis is the definition 4 rational ........i think so
anonymous
  • anonymous
Yes, its okay, it was just a little slip
anonymous
  • anonymous
He meant the other way round
anonymous
  • anonymous
squaring both sides makes the equation quadratic..........which gives an xtra solution......
anonymous
  • anonymous
Which equation are you talking of
anonymous
  • anonymous
u may take any
anonymous
  • anonymous
say x=5
anonymous
  • anonymous
squaring both sides ......... x^2=25 x=5 or -5
anonymous
  • anonymous
bt x=5
anonymous
  • 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
  • anonymous
ahmm...thnx aditya frm san diego......ryt!!!!!!!!!
anonymous
  • anonymous
I concluded that always I will have to go back to the original equation, and plug in my solutions
anonymous
  • anonymous
jas think b4 squarng
anonymous
  • anonymous
But its hard to accept, although its the truth, as I never did it, all these days.
anonymous
  • 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
  • anonymous
Very true, but still a bad news for me
anonymous
  • anonymous
Anyhow thank you all
anonymous
  • 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
  • anonymous
Well, lets come back to the problem I began with
anonymous
  • anonymous
I was trying to prove that \[\sqrt{2}+\sqrt5\]
anonymous
  • anonymous
I am trying to prove that its a irrational number
anonymous
  • anonymous
Lets use the following reasoning \[[x-\sqrt (2)]^2=5\]
anonymous
  • anonymous
\[x^2-3=2\sqrt2 x\]
anonymous
  • anonymous
\[\sqrt(2)=(x^2-3)(2x)\]
anonymous
  • anonymous
So if I assumed x to be a rational number, then here we get a contradiction
anonymous
  • 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
  • anonymous
I mean here, its no point plugging back, then what is the fact that allows me to rely on the result
anonymous
  • anonymous
You introduced quite a lot without telling us where it came from (x- sq rt 2)^2=5?
anonymous
  • anonymous
I squared both sides
anonymous
  • anonymous
First I assumed √2+√5=x where x=p/q
anonymous
  • anonymous
Where p and q are co primes
anonymous
  • anonymous
Finally I proved that by that assumption we meet at a contradiction
anonymous
  • anonymous
So I commented that √2+√5 can't be a rational number
anonymous
  • 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
  • anonymous
Are you getting my question?
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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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