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Solve radical equation

  1. Feb 3, 2008 #1
    Could a tutor check my solution to my radical question? It is enclosed in this attachment as a word document.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Feb 3, 2008 #2
    Your attachment needs to be approved. Either way, it's still pretty dangerous to open word attachments.

    Type it up and we'll take a look at it.
     
  4. Feb 3, 2008 #3

    HallsofIvy

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    Why don't check your own answer? Just put your answer back into the original equation and do the arithmetic.
     
  5. Feb 3, 2008 #4
    You can plug your answer back into the equation to see if it works out. Its absolutely correct & absolutely wrong all at the same time. Can you figure out why? Plug your values back in & see why I said what I said. Your first step contains an algebra error, also think factoring for the first step.
     
  6. Feb 5, 2008 #5
    Is this correct?

    3x + x*sqrt(5) = 2

    solution

    (3x + x*sqrt(5)) = 2

    I factored out the x
    x (3 + sqrt(5)) = 2

    Divide both sides by sqrt(5) + 3
    x (3 + sqrt(5) / (sqrt(5) + 3) = (2) / (sqrt(5) + 3)

    cancel

    x = (2) / (sqrt(5) + 3) final answer

    Is this correct?
     
    Last edited: Feb 5, 2008
  7. Feb 5, 2008 #6
    3x + x√5 = 2

    ²√5 = 5^½
    3x + (5^½)x = 2
    ~5.24x=2
    x=~0.38

    i get the same
     
    Last edited: Feb 5, 2008
  8. Feb 6, 2008 #7

    HallsofIvy

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    Yes, that is correct. You could make it look nicer by "rationalizing the denominator": multiply both numerator and denominator by [itex]\sqrt{5}- 3[/itex] and you get
    [tex]x= \frac{2\sqrt{5}- 6}{2}= \sqrt{5}- 3[/itex]

    By the way, none of these are really "radical" equations since they don't involve roots of the unknown number. [itex]\sqrt{5}[/itex] is just another number!

    Oh, and there are no "tutors" here. Just us folks.
     
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