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ligia587

which is expression is equivalent to

  • one year ago
  • one year ago

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  1. ligia587
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    if \[\sqrt{x-a}=b, x>a \]

    • one year ago
  2. ligia587
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    which expression is equivalent to x? \[b^{2}-a \]

    • one year ago
  3. ligia587
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    \[b-a\]

    • one year ago
  4. ligia587
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    b+a

    • one year ago
  5. ligia587
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    or b^2+a

    • one year ago
  6. kirbykirby
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    \[\sqrt{x-a}=b\]=>\[x-a=b^2\]=>\[x=b^2+a\]

    • one year ago
  7. kirbykirby
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    When solving with radicals, square both sides because \[(\sqrt{x})^2=x\]

    • one year ago
  8. kirbykirby
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    It will just get rid of it :)!!

    • one year ago
  9. ligia587
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    so b^2+ a is the answer?

    • one year ago
  10. kirbykirby
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    yes

    • one year ago
  11. ligia587
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    how did you know x=> ?

    • one year ago
  12. kirbykirby
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    Did you understand the first step? that x-a = b^2 ?

    • one year ago
  13. ligia587
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    no i dont

    • one year ago
  14. ligia587
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    |dw:1358391793350:dw|

    • one year ago
  15. kirbykirby
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    Ok. Do you understand why \[(\sqrt{x})^2=x\]?

    • one year ago
  16. kirbykirby
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    The "x > a" is just imposing a restriction value on your x. You don't need to use it to solve for x

    • one year ago
  17. ligia587
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    where did you get the square from?

    • one year ago
  18. kirbykirby
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    It's what you need to do in order to solve the equation. It's similar to like when you have an equation like: \[2x-4=10\]\[2x=10+4\]\[2x=14\]\[\frac{2x}{2}=\frac{14}{2}\]\[x=7\] You could argue "how did you get the divided by 2 part?" Well it's similar to your situation. You introduce this exponent to get rid of your square root, just like you introduced a divided by 2 up here^ to isolate your x

    • one year ago
  19. ligia587
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    ohh ok

    • one year ago
  20. ligia587
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    thanks so much :)

    • one year ago
  21. kirbykirby
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    No problem :)

    • one year ago
  22. ligia587
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    can you help me with another problem?

    • one year ago
  23. kirbykirby
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    SUre

    • one year ago
  24. kirbykirby
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    @ligia587

    • one year ago
  25. ligia587
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    yea?

    • one year ago
  26. ligia587
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    i thought you left

    • one year ago
  27. kirbykirby
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    I can help your with your question if you need

    • one year ago
  28. kirbykirby
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    you*

    • one year ago
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