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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    I can't think of any restrictions on squaring both sides of an equation.

  2. anonymous
    • 5 years ago
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    Can you square the equation |3|=x I think if you do that you will loose solutions.

  3. anonymous
    • 5 years ago
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    yes, you are right.

  4. anonymous
    • 5 years ago
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    But I know that that knowledge is not enough. Can you please provide me some more of such things

  5. anonymous
    • 5 years ago
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    I mean a little details, and when not to square etc. etc

  6. anonymous
    • 5 years ago
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    I would, but I am very sleepy now. I'll think of some and come back. This is interesting stuff.

  7. anonymous
    • 5 years ago
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    Ok thank you

  8. anonymous
    • 5 years ago
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    you should ask INewton and Lokisan. I think they will be able to help you the best.

  9. anonymous
    • 5 years ago
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    Ok I will do that

  10. anonymous
    • 5 years ago
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    Glad that you are here

  11. anonymous
    • 5 years ago
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    Could you help please

  12. anonymous
    • 5 years ago
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    There are no restrictions.

  13. anonymous
    • 5 years ago
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    What is causing you to think that?

  14. anonymous
    • 5 years ago
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    You see, I often get more solutions than what I am expected to get, when I square both sides

  15. anonymous
    • 5 years ago
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    For example If I start with sqrt 2 + sqrt 5 =x sqrt 2 =x - sqrt 5

  16. anonymous
    • 5 years ago
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    and then square both sides and go on, I will end up with sqrt 2 +- sqrt 5

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

  18. anonymous
    • 5 years ago
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    So now if we use the formula for quadratic equation

  19. anonymous
    • 5 years ago
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    We get two things

  20. anonymous
    • 5 years ago
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    right?

  21. anonymous
    • 5 years ago
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    Yes

  22. anonymous
    • 5 years ago
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    You're getting two answers now because you've asked a different question.

  23. anonymous
    • 5 years ago
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    "different question"?

  24. anonymous
    • 5 years ago
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    Solving your first equation is not the same as solving the squared one.

  25. anonymous
    • 5 years ago
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    yes, i think lokisan is right. you asked a different question when you take square root on both sides

  26. anonymous
    • 5 years ago
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    I see what you're thinking, but you needn't worry :)

  27. anonymous
    • 5 years ago
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    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,

  28. anonymous
    • 5 years ago
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    Ok

  29. anonymous
    • 5 years ago
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    So does that mean I shouldn't square any equation

  30. anonymous
    • 5 years ago
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    I know you're not convinced, but just think about it for a while.

  31. anonymous
    • 5 years ago
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    :)

  32. anonymous
    • 5 years ago
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    Why did you think you had to square it anyway?

  33. anonymous
    • 5 years ago
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    lol

  34. anonymous
    • 5 years ago
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    Just working with irrational numbers

  35. anonymous
    • 5 years ago
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    I was trying to prove that the sum of these two irrational numbers will lead to one irrational number

  36. anonymous
    • 5 years ago
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    You can't get rid of irrational sums by squaring!

  37. anonymous
    • 5 years ago
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    Oh...see above.

  38. anonymous
    • 5 years ago
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    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

  39. anonymous
    • 5 years ago
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    This made me question, "what are the restrictions..........."

  40. anonymous
    • 5 years ago
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    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.

  41. anonymous
    • 5 years ago
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    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

  42. anonymous
    • 5 years ago
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    But if you try squaring them, you will find that we get a solution

  43. anonymous
    • 5 years ago
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    And you will have to square twice

  44. anonymous
    • 5 years ago
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    nope, wolfram alfa says no solutions exist.

  45. anonymous
    • 5 years ago
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    If you have time, then I can square both sides, and solve it here

  46. anonymous
    • 5 years ago
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    I will do it manually, on paper.

  47. anonymous
    • 5 years ago
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    Yes, that is what I am talking of

  48. anonymous
    • 5 years ago
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    yes, so solve it here, iamignorant/

  49. anonymous
    • 5 years ago
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    sqrt(x+1)-sqrt(x-1)=sqrt(4x-1)

  50. anonymous
    • 5 years ago
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    x+1+x-1-2sqrt(x+1)(x-1)=4x-1

  51. anonymous
    • 5 years ago
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    2x-2sqrt(x+1)(x-1)=4x-1

  52. anonymous
    • 5 years ago
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    -2sqrt(x+1)(x-1)=2x-1

  53. anonymous
    • 5 years ago
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    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.

  54. anonymous
    • 5 years ago
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    Yes exactly

  55. anonymous
    • 5 years ago
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    But how do I know

  56. anonymous
    • 5 years ago
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    I mean will I have to plug in every value after solving every equation of the same type

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

  58. anonymous
    • 5 years ago
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    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.

  59. anonymous
    • 5 years ago
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    Just what I said.

  60. anonymous
    • 5 years ago
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    Does that mean I must always check back?

  61. anonymous
    • 5 years ago
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    Yes.

  62. anonymous
    • 5 years ago
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    Now what to say......

  63. anonymous
    • 5 years ago
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    How about, "I get it"?

  64. anonymous
    • 5 years ago
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    sorry?

  65. anonymous
    • 5 years ago
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    No...you just learnt something.

  66. anonymous
    • 5 years ago
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    I am sorry, but I am still not getting you

  67. anonymous
    • 5 years ago
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    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...)

  68. anonymous
    • 5 years ago
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    Well, it's late, so I'm not really bothered with semantics.

  69. anonymous
    • 5 years ago
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    Are you satisfied, Iam?

  70. anonymous
    • 5 years ago
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    I will have to be, no other option

  71. anonymous
    • 5 years ago
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    I have to go - just think about it...

  72. anonymous
    • 5 years ago
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    :Sure I will

  73. anonymous
    • 5 years ago
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    Geeeeeez that guy's a buzzkill.

  74. anonymous
    • 5 years ago
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    Whatever, he has helped me a lot since the time he joined openstudy

  75. anonymous
    • 5 years ago
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    And also is a smart man

  76. anonymous
    • 5 years ago
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    That doesn't mean I am trying to offend you Mr Newton

  77. anonymous
    • 5 years ago
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    My comment was somewhat tongue-in-cheek. But, how smart?

  78. anonymous
    • 5 years ago
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    Actually I am ignorant, and any comment from my side will simply be a ridiculous one

  79. anonymous
    • 5 years ago
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    :)

  80. anonymous
    • 5 years ago
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    Any how

  81. anonymous
    • 5 years ago
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    Mr Newton, this thing of plugging final solutions, and checking them

  82. anonymous
    • 5 years ago
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    What about it?

  83. anonymous
    • 5 years ago
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    Should it be done always, or only when we are dealing with square roots

  84. anonymous
    • 5 years ago
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    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)

  85. anonymous
    • 5 years ago
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    You mean, when I am searching for roots, only then I need to actually do these things

  86. anonymous
    • 5 years ago
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    right?

  87. anonymous
    • 5 years ago
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    I meant square roots. Roots of the equation would not (normally) have to be checked.

  88. anonymous
    • 5 years ago
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    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

  89. anonymous
    • 5 years ago
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    This was the question

  90. anonymous
    • 5 years ago
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    If I start with sqrt 2 + sqrt 5 =x sqrt 2 =x - sqrt 5

  91. anonymous
    • 5 years ago
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    I squared both sides and then I solved it

  92. anonymous
    • 5 years ago
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    I ended up with sqrt 2 +- sqrt 5

  93. anonymous
    • 5 years ago
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    So Mr Lucas commented, that I actually changed the question by squaring it

  94. anonymous
    • 5 years ago
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    Could you please provide me a little more explanation

  95. anonymous
    • 5 years ago
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    I mean where I changed it, and how did you all guess, that I changed it

  96. anonymous
    • 5 years ago
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    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.

  97. anonymous
    • 5 years ago
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    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.

  98. anonymous
    • 5 years ago
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    Have a nice day.

  99. anonymous
    • 5 years ago
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    Yes got you

  100. anonymous
    • 5 years ago
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    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.

  101. anonymous
    • 5 years ago
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    dis is the definition 4 rational ........i think so

  102. anonymous
    • 5 years ago
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    Yes, its okay, it was just a little slip

  103. anonymous
    • 5 years ago
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    He meant the other way round

  104. anonymous
    • 5 years ago
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    squaring both sides makes the equation quadratic..........which gives an xtra solution......

  105. anonymous
    • 5 years ago
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    Which equation are you talking of

  106. anonymous
    • 5 years ago
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    u may take any

  107. anonymous
    • 5 years ago
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    say x=5

  108. anonymous
    • 5 years ago
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    squaring both sides ......... x^2=25 x=5 or -5

  109. anonymous
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    bt x=5

  110. anonymous
    • 5 years ago
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    yes, siddharth is right. Squaring both sides makes it a quadratic equation, providing an extra or as Inewton called it, a spurious root.

  111. anonymous
    • 5 years ago
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    ahmm...thnx aditya frm san diego......ryt!!!!!!!!!

  112. anonymous
    • 5 years ago
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    I concluded that always I will have to go back to the original equation, and plug in my solutions

  113. anonymous
    • 5 years ago
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    jas think b4 squarng

  114. anonymous
    • 5 years ago
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    But its hard to accept, although its the truth, as I never did it, all these days.

  115. anonymous
    • 5 years ago
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    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.

  116. anonymous
    • 5 years ago
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    Very true, but still a bad news for me

  117. anonymous
    • 5 years ago
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    Anyhow thank you all

  118. anonymous
    • 5 years ago
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    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.

  119. anonymous
    • 5 years ago
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    Well, lets come back to the problem I began with

  120. anonymous
    • 5 years ago
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    I was trying to prove that \[\sqrt{2}+\sqrt5\]

  121. anonymous
    • 5 years ago
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    I am trying to prove that its a irrational number

  122. anonymous
    • 5 years ago
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    Lets use the following reasoning \[[x-\sqrt (2)]^2=5\]

  123. anonymous
    • 5 years ago
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    \[x^2-3=2\sqrt2 x\]

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

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

  126. anonymous
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    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.

  127. anonymous
    • 5 years ago
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    I mean here, its no point plugging back, then what is the fact that allows me to rely on the result

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

  129. anonymous
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    I squared both sides

  130. anonymous
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    First I assumed √2+√5=x where x=p/q

  131. anonymous
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    Where p and q are co primes

  132. anonymous
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    Finally I proved that by that assumption we meet at a contradiction

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

  134. anonymous
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    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

  135. anonymous
    • 5 years ago
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    Are you getting my question?

  136. anonymous
    • 5 years ago
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    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 :(

  137. anonymous
    • 5 years ago
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    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

  138. anonymous
    • 5 years ago
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    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

  139. anonymous
    • 5 years ago
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    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

  140. anonymous
    • 5 years ago
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    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).

  141. anonymous
    • 5 years ago
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    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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spraguer (Moderator)
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