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ganeshie8

i am getting two solutions for this problem. but only one is satisfying the equation. could someone pls explain whats going on... http://assets.openstudy.com/updates/attachments/50064e0ae4b0624180677e0a-emily9102-1342590484915-mathproblem5.jpg

  • one year ago
  • one year ago

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  1. f1r1e1s1h1
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    square it and try?

    • one year ago
  2. ganeshie8
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    |dw:1342591850277:dw|

    • one year ago
  3. ganeshie8
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    x =1 is not satisfying the equation :( not sure why

    • one year ago
  4. ganeshie8
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    hmm

    • one year ago
  5. f1r1e1s1h1
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    man u have squared it so ofcourse one will not match?

    • one year ago
  6. higgs
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    when you square something, you will always get two solutions, and not all of them will be right. substitute the answers back into the original equation to check. in this case, 1 is not the solution as -1 is not equal to 1

    • one year ago
  7. ganeshie8
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    im li'l lost.... could u elaborate pls.. ?

    • one year ago
  8. ganeshie8
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    i see we are squaring... and losing sign sensitivity is it

    • one year ago
  9. mukushla
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    u must consider restrictions in every step note that when u have \( x-3=2 \sqrt{x} \) then \( x-3 \ge0 \)

    • one year ago
  10. ganeshie8
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    ah that makes sense. thanks mukushla straight to the point. :)

    • one year ago
  11. mukushla
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    welcome my friend

    • one year ago
  12. higgs
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    I suppose this also applies to irrational equations like this? \[\sqrt{x-1}+\sqrt{x+4}=\sqrt{3x+10}\] I got \(x = 5\) and \(x = \frac{-13}{3} \) \(x = \frac{-13}{3} \) is rejected in this case as well because after substituting into the equation, it became undefined.

    • one year ago
  13. ganeshie8
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    yeah now i understand thanks higgs :D here, from the first equation itself, we can apply the restriction x >=1. and hence rejected solutions < 1

    • one year ago
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