Question

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

Answer 1

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

Answer 2

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

Answer 3

yes, you are right.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Ok thank you

Answer 8

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

Answer 9

Ok I will do that

Answer 10

Glad that you are here

Answer 11

Could you help please

Answer 12

There are no restrictions.

Answer 13

What is causing you to think that?

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

So now if we use the formula for quadratic equation

Answer 19

We get two things

Answer 20

right?

Answer 21

Yes

Answer 22

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

Answer 23

"different question"?

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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,

Answer 28

Ok

Answer 29

So does that mean I shouldn't square any equation

Answer 30

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

Answer 31

:)

Answer 32

Why did you think you had to square it anyway?

Answer 33

lol

Answer 34

Just working with irrational numbers

Answer 35

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

Answer 36

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

Answer 37

Oh...see above.

Answer 38

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

Answer 39

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

Answer 40

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.

Answer 41

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

Answer 42

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

Answer 43

And you will have to square twice

Answer 44

nope, wolfram alfa says no solutions exist.

Answer 45

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

Answer 46

I will do it manually, on paper.

Answer 47

Yes, that is what I am talking of

Answer 48

yes, so solve it here, iamignorant/

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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.

Answer 54

Yes exactly

Answer 55

But how do I know

Answer 56

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

Answer 57

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

Answer 58

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.

Answer 59

Just what I said.

Answer 60

Does that mean I must always check back?

Answer 61

Yes.

Answer 62

Now what to say......

Answer 63

How about, "I get it"?

Answer 64

sorry?

Answer 65

No...you just learnt something.

Answer 66

I am sorry, but I am still not getting you

Answer 67

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...)

Answer 68

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

Answer 69

Are you satisfied, Iam?

Answer 70

I will have to be, no other option

Answer 71

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

Answer 72

:Sure I will

Answer 73

Geeeeeez that guy's a buzzkill.

Answer 74

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

Answer 75

And also is a smart man

Answer 76

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

Answer 77

My comment was somewhat tongue-in-cheek.

But, how smart?

Answer 78

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

Answer 79

:)

Answer 80

Any how

Answer 81

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

Answer 82

What about it?

Answer 83

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

Answer 84

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)

Answer 85

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

Answer 86

right?

Answer 87

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

Answer 88

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

Answer 89

This was the question

Answer 90

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

Answer 91

I squared both sides and then I solved it

Answer 92

I ended up with sqrt 2 +- sqrt 5

Answer 93

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

Answer 94

Could you please provide me a little more explanation

Answer 95

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

Answer 96

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.

Answer 97

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.

Answer 98

Have a nice day.