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Homework Help: How to expand a radical

  1. Mar 25, 2009 #1
    1. The problem statement, all variables and given/known data
    How do you expand a radical?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 25, 2009 #2
    You have to rephrase your question. As far as I know, you can only expand a term, e.g. [tex] (x - 2)^2 [/tex]
  4. Mar 25, 2009 #3
    Sorry I meant a term. like you example you just gave. I am confused on how to expand them for doing radical equations.
  5. Mar 25, 2009 #4
    [tex] (x - \sqrt{2})^2 = (x - \sqrt{2})(x - \sqrt{2}) [/tex]

    Do you know how to F.O.I.L.? (multiply first by outer and inner by last)

    [tex] x^2- x\sqrt{2} - x\sqrt{2} + 2 [/tex]


    [tex] (a + b)^2 = a^2 + 2ab + b^2 [/tex]

    Just take it slow. You can always check your work with a calculator.
  6. Mar 25, 2009 #5
    Oh so you multiply from inner to outer. I get it now.
  7. Mar 26, 2009 #6

    I am having trouble with this problem, I used the quadratic formual and the binomial square formula. I got x=9.75561, I just checked it and its wrong. I think the problem is when i try to exand [tex]\sqrt{2x*2}[/tex], after the expantion of the right side i got 484-4[tex]^{x}_{2}[/tex]-8x+4.
  8. Mar 26, 2009 #7


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    Science Advisor

    [itex]\sqrt{2x*2}[/itex]? That's just [itex]2\sqrt{x}[/itex]. Do you mean [itex]\sqrt{2x+2}[/itex]?
  9. Mar 26, 2009 #8
    I'm having trouble understanding your notation. Is that a 2x TIMES 2 inside the first square root? Can you clean that up?

  10. Mar 26, 2009 #9
    Sorry, I made a typo it was suppose to be +.
  11. Mar 26, 2009 #10
    You can't expand [tex] \sqrt{2x + 2} [/tex], because it isn't a term raised to a power. You understand me?

    Say you had [tex] (\sqrt{2x + 2})^2 [/tex], you could expand that, because it's a term raised to a power.

    If your question is [tex] \sqrt{2x + 2} + \sqrt{3x} = 22 [/tex] , then try squaring both sides, and see what happens.
  12. Mar 26, 2009 #11
    If I only square each side and leave the to square roots on the left alone the when I comebine the 3x and the 2x i get 5x+2=484, then i sybract the 2 to get 5x=482, I then do the divition and I get 96.4. That's not right.
  13. Mar 26, 2009 #12
    Oh, yeah, move the [tex] \sqrt{3x} [/tex] to the right hand side. Then square each side.
  14. Mar 26, 2009 #13
    You're going to have the term [tex] (22 - \sqrt{3x}) [/tex] to expand on the RHS.
  15. Mar 27, 2009 #14


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    Homework Helper

    No sorry, I don't understand you..

    [tex]\sqrt{a}=a^{\frac{1}{2}}[/tex] and that's a term raised to a power. It's just not an integral power.

  16. Mar 27, 2009 #15
    Can you expand [tex]\sqrt{2x+2}=(2x+2)^{\frac{1}{2}}[/tex] ? I'm sorry I wasn't more accurate, but I was just trying to help this person with their homework. If by some chance I came off as snobby(or whatever), I didn't intend for it.
  17. Mar 27, 2009 #16
    You're right, that's incorrect. So you tried to do
    [tex](\sqrt{2x + 2} + \sqrt{3x})^{2}[/tex]
    and you got
    [tex]2x + 2 + 3x[/tex]
    which is incorrect. Squaring that radical expression doesn't work that way.

    Try what General Sax suggested, that is, first subtract the [tex]\sqrt{3x}[/tex] from both sides to get
    [tex]\sqrt{2x + 2} = 22 - \sqrt{3x}[/tex]

    Now square both sides.
    [tex](\sqrt{2x + 2})^{2} = (22 - \sqrt{3x})^{2}[/tex]
    In this case the left side becomes 2x + 2. You'll need to use the special product property [tex](a - b)^{2}=a^{2} - 2ab + b^{2}[/tex] to expand the right side. Can you take it from there?

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