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Homework Help: Square roots question

  1. Jan 5, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\frac{5}{\sqrt{7+3\sqrt{x}}}[/tex] = [tex]\sqrt{7 -3\sqrt{x}}[/tex]

    2. Relevant equations


    3. The attempt at a solution
    does this equal 5 = 7 - 3[tex]\sqrt{x}[/tex]
  2. jcsd
  3. Jan 5, 2008 #2
    i cross multiplied, but i'm a little iffy on my FOIL technique when square roots are throw into the mix
  4. Jan 5, 2008 #3


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    First square both sides and then cross multiply. That's what i'd do.
  5. Jan 5, 2008 #4
    5 = sqrt (49 - 9x - 42 sqrt x)
    25 = 49 - 9x - 42 sqrt x
    24 = 9x + 42 sqrt x
    The rest I leave it to you. You just have to square another time.
  6. Jan 5, 2008 #5
    i don't understand how you got that could you explain it please?
  7. Jan 5, 2008 #6


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    This term shouldn't be there. Check your signs.
    Last edited: Jan 5, 2008
  8. Jan 5, 2008 #7
    okay so square both sides to get rid of the first square root sign then square again to get rid of the one on the "x" ?
  9. Jan 5, 2008 #8


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    Square both sides (this gets rid of the "first set" of square roots). Then cross multiply. Expand the right hand side. Note your laws of exponents: so [tex] x^\frac{1}{2} * x^\frac{1}{2} = x ^ { \frac{1}{2} + \frac{1}{2}} = x^1 = x[/tex]
  10. Jan 5, 2008 #9
    okay i'm definately lost now.

    i have the stated question

    i sqaure both sides, i get

    [tex]\frac{25}{7 + 2((\sqrt{7})(\sqrt{3\sqrt{x}})) + 3\sqrt{x}}[/tex] = 7 + 2(([tex]\sqrt{7}[/tex])([tex]\sqrt{-3\sqrt{x}}[/tex])) + 3[tex]\sqrt{x}[/tex]
  11. Jan 5, 2008 #10
    how can i cross multiply that without getting utterly lost????
  12. Jan 5, 2008 #11


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    By square both sides, I mean

    [tex](\frac{5}{\sqrt{7+3\sqrt{x}}})^2[/tex] = [tex](\sqrt{7 -3\sqrt{x}} )^2[/tex]

    This means squaring every term

    [tex]\frac{5^2}{(\sqrt{7+3\sqrt{x}})^2}[/tex] = [tex](\sqrt{7 -3\sqrt{x}} )^2[/tex]

    Can you tell me what this is equivalent to: [tex](\sqrt{x})^2[/tex]? If you know this, then you will know how to make the above much easier to work with. Figure this out before you expand anything. See my laws of exponents remark.

    By the way the curved brackets in the second equation go around the whole square root term. Hopefully it doesn't look too confusing to you.
    Last edited: Jan 5, 2008
  13. Jan 5, 2008 #12
    sqrt x squared is just X. but i get confused with the middle terms. im going to try it later on when i have the time to, i'll get back to you
  14. Jan 5, 2008 #13


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    Yes, that's right. So if you have

    [tex](\sqrt{7 -3\sqrt{x}} )^2[/tex]

    and apply that idea, what does this term become?
  15. Jan 5, 2008 #14


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    The parentheses are peculiarly placed!
    [tex]\frac{5}{\sqrt{7+3\sqrt{x}}}= \sqrt{7 -3\sqrt{x}}[/tex]
    on both sides by [itex]\sqrt{7+ 3\sqrt{x}}[/itex] gives [itex]5= \sqrt{(7- 3\sqrt{x})(7+ 3\sqrt{x})}= \sqrt{49- 9x}[/itex], because the product inside the squareroot is "sum times difference". Now squaring gives 25= 49- 9x. That should be easy to solve. Don't forget to check your answer- multiplying both sides by something involving x or squaring both sides can introduce "extraneous" solutions.
  16. Jan 5, 2008 #15
    c=8 i think
  17. Jan 6, 2008 #16


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    x = 8/3 I think.
  18. Jan 6, 2008 #17


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    Since there isno "c" in the problem, I guess you missed the "x" key!
    Unfortunately, with x= 8, the [itex]7- 3\sqrt{8}[/itex] is clearly negative and so cannot satisfy the equation.

    Check it!
    [tex]\frac{5}{\sqrt{7+ 3\sqrt{8/3)}}}= 1.4494897427831780981972840747059[/tex]
    [tex]\sqrt{7- 3\sqrt{(8/3)}}= 1.4494897427831780981972840747059[/tex]
    Looks good to me!

    As I said before,
    If 25= 49- 9x, the 9x= 49-25= 24 so x= 24/9= 8/3 as epenguin said. (But you should still check numerically.)
  19. Jan 6, 2008 #18
    lol i just rechecked my work and i'm an idiot. i had 24/9 but i divded both by 3 but the i just decided not to include the /3 part. lol @ me thanks for the help tho
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