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Equations Involving Radicals Question

  1. Oct 15, 2012 #1
    Solve for:
    √(x-7) / √(x) -2 = √2

    My attempt at a solution:
    I solved for x and it comes out to:

    0 = x^2 - 64x +225

    and then i plugged it into the quadratic formula:

    [-(-64)±√((64)^2-4(1)(225))]/2

    and my answer comes out to be:

    32±√799

    although the answer on the back of the package is 9 and 25.

    kind of confused on how to attempt this question now and not sure what i'm doing wrong...

    And thanks in advance for the help!!
     
  2. jcsd
  3. Oct 15, 2012 #2

    Dick

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    0 = x^2 - 64x +225 isn't the right quadratic. Check it again or show how you got it.
     
  4. Oct 15, 2012 #3
    (√(x-7)) / (√(x) - 2) = √2

    √(x-7) = √(2) (√(x) - 2)

    x-7 = (√2x - 2√2) ^2 <------- square both sides

    x-7 = (√2x)^2 - 2(√2x)(2√2) + (√8)^2

    x-7 = 2x - 8√x + 8

    0 = x - 8√x +15

    0 = x^2 - 64x + 225 <----- squared both sides to get rid of radical

    and plug into quadratic eqn. :P
     
  5. Oct 15, 2012 #4

    SammyS

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    Is the problem your working on,

    [itex]\displaystyle \frac{\sqrt{x-7\,}}{\sqrt{x\,}-2}=\sqrt{2} \ ?[/itex]

    Or is it [itex]\displaystyle \frac{\sqrt{x-7\,}}{\sqrt{x\,}}-2=\sqrt{2} \ ?[/itex] which is literally what you wrote.
     
  6. Oct 15, 2012 #5
    the first one
    Sorry i don't really know how to use the square root sign, pretty new to this
     
  7. Oct 15, 2012 #6

    Dick

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    You were doing fine till you tried to square both sides of 0 = x - 8√x +15. Squaring both sides won't get rid of the radical if you do the algebra right. Best to write it as 8√x = x+15 first. Now square both sides.
     
  8. Oct 15, 2012 #7
    OHHHHH That make sense!!

    thanks!!
     
  9. Oct 15, 2012 #8

    SammyS

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    It's more a problem of leaving out or of misplacing parentheses.

    Writing , √(x-7) / ( √(x) -2 ) = √2 would be OK .
     
  10. Oct 15, 2012 #9

    Ray Vickson

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    Square toot signs are not the issue; brackets are what you missed. Everything would have been clear if you had written (√(x-7)) / (√(x) - 2) = √2 . The point is that writing something like A/B-C means (A/B) - C if you read it using standard priority rules for mathematical expressions. If you want A/(B-C), you need brackets.

    RGV
     
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