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airyana1114

  • one year ago

Can someone help me solve for x?

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  1. airyana1114
    • one year ago
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    \[2\sqrt{x-5}=2\]

  2. airyana1114
    • one year ago
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    I know my first step is to subtract 2 from both sides of the equation which gives me... \[\sqrt{x-5}=0\] Then I subtract 5 from both sides and I'm left with... \[\sqrt{x}=5\] Is there another step after that?

  3. zepdrix
    • one year ago
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    Wooooops! :O

  4. zepdrix
    • one year ago
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    If the 2 was being `added` to the root, then yes, you would `subtract it from both sides to undo the addition.

  5. zepdrix
    • one year ago
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    But in this case, it looks like the 2 is `multiplying` the root, yes?

  6. zepdrix
    • one year ago
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    So how do we undo multiplication? :)

  7. airyana1114
    • one year ago
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    We divide :D

  8. zepdrix
    • one year ago
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    Good, yes :) Do that. It should give you something a little different than 0 on the right side.

  9. airyana1114
    • one year ago
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    Okay, so now I got 1 instead of 0. @zepdrix

  10. zepdrix
    • one year ago
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    \[\large\rm \sqrt{x-5}=1\]Cool.

  11. airyana1114
    • one year ago
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    Then I would add 5 and get \[\sqrt{x}=6\] @zepdrix

  12. zepdrix
    • one year ago
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    Wooops! The subtraction is `inside` of the root. We can't do that operation yet. We need to deal with the root first.

  13. zepdrix
    • one year ago
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    What's the inverse of `square root`? How do we undo that? :o

  14. airyana1114
    • one year ago
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    Okay so square both sides of the equation? It would still be 1 correct?

  15. zepdrix
    • one year ago
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    \[\large\rm \left(\sqrt{x-5}\right)^2=1^2\]On the left, the operations undo one another, good. On the right, 1^2 = 1? Sounds good!\[\large\rm x-5=1\]

  16. airyana1114
    • one year ago
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    Okay, then I would add 5 to both sides of the equation correct? @zepdrix

  17. anonymous
    • one year ago
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    yes

  18. airyana1114
    • one year ago
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    Thank you :) @satellite73

  19. zepdrix
    • one year ago
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    yay good job \c:/

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