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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
I'm not following. If you square both sides, you should get\[(\sqrt{2})^2=(x-\sqrt{5})^2 \rightarrow 2=x^2-2\sqrt{5}x+5\]
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.
yes, i think lokisan is right. you asked a different question when you take square root on both sides
I see what you're thinking, but you needn't worry :)
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,
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
I was trying to prove that the sum of these two irrational numbers will lead to one irrational number
You can't get rid of irrational sums by squaring!
Oh...see above.
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
This made me question, "what are the restrictions..........."
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.
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
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
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.
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
You ended up at a contradiction assuming this thing had a solution. The contradiction shows the assumption to be false (reductio ad absurdum).
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.
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
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...)
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
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)
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.
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
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
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.
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.
Have a nice day.
Yes got you
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.
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
yes, siddharth is right. Squaring both sides makes it a quadratic equation, providing an extra or as Inewton called it, a spurious root.
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.
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.
Very true, but still a bad news for me
Anyhow thank you all
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.
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
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.
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
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
Are you getting my question?
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 :(
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
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
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
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).
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question