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ligia587

  • one year ago

which is expression is equivalent to

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

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

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

  4. ligia587
    • one year ago
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    b+a

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

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

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

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

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

  10. kirbykirby
    • one year ago
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    yes

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

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

  13. ligia587
    • one year ago
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    no i dont

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

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

  16. kirbykirby
    • one year ago
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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

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

  18. kirbykirby
    • one year ago
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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

  19. ligia587
    • one year ago
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    ohh ok

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

  21. kirbykirby
    • one year ago
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    No problem :)

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

  23. kirbykirby
    • one year ago
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    SUre

  24. kirbykirby
    • one year ago
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    @ligia587

  25. ligia587
    • one year ago
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    yea?

  26. ligia587
    • one year ago
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    i thought you left

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

  28. kirbykirby
    • one year ago
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    you*

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