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Help with a power equation

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data
    (5R2+11)1/2=-4R

    (It's actually written as a squareroot rather than being raised to the 1/2 power, but couldn't get latex to do it's thing.)


    2. Relevant equations
    None?


    3. The attempt at a solution
    I tried a couple different ways, none yeilding the correct answer. eg:
    Squaring both sides to get
    5R2+11=16R2

    Combining like terms (16R2-5R2) to get
    11=11R2

    Without going any further, I can already tell that this is the wrong answer. Continuing, I'll get 1 rather than -1 (the solution when I solve graphically, also in the back of the book)

    I also tried to begin by dividing both sides by -4. That didn't really get me anywhere.

    Edit: I thought I should elaborate on that...
    Dividing by -4
    [(5R2+11)1/2]/-4=R

    Getting R by itself doesn't really change anything for me. I still need to "get rid" of the square root. It also gives me an unpleasant fraction to work with. I'd rather not go that route if I don't have to.

    I've got some kind of mental block that won't let me see what I'm doing wrong. You don't have to give me all the steps to solve the problem, maybe just tell me what I did wrong and point me in the right direction.

    Thanks in advance.
     
    Last edited: Sep 24, 2011
  2. jcsd
  3. Sep 24, 2011 #2
    [itex]\sqrt{5x^2+11}=-4x[/itex]

    [itex]5x^2+11=16x^2[/itex]

    [itex]11=11x^2[/itex]

    [itex]\sqrt{1}=x[/itex]

    [itex]\pm1=x[/itex]

    ^ Are you sure that's not the answer in your book?
     
  4. Sep 24, 2011 #3
    I missed a vital step.

    I know what x equals, so now running the original equation with the values of x, I can see that x can only be -1.

    If it's a positive one, then you're stating that you've taken a square root and returned a negative value.

    Thus:

    [itex]x = -1[/itex]
     
  5. Sep 24, 2011 #4
    Thanks AJKing! I made a stupid mistake. [itex]\sqrt{1}[/itex] should be [itex]\pm1[/itex]. My failure to solve this problem makes much more sense now that you've pointed this out to me. I forgot the negative solution...

    I haven't run into a situation quite like this one before where only one of the solutions is a solution. (You guys can help me with the vocab there.) Is there a way to know (while solving the problem as opposed to checking) that only one answer is a solution?
     
    Last edited: Sep 24, 2011
  6. Sep 24, 2011 #5

    Mark44

    Staff: Mentor

    The line above should be [itex]\pm \sqrt{1} = x[/itex]

    [itex]\sqrt{1} = 1[/itex]
    The square root of 1 has one value, contrary to what many people mistakenly think.

     
  7. Sep 24, 2011 #6
    Wait, what? I thought (-1)2=1. If [itex]\sqrt{1}[/itex] doesn't have a negative solution, how can the solution to my original equation be negative?

    I'm lost now.
     
  8. Sep 24, 2011 #7

    Mark44

    Staff: Mentor

    Yes, this is true.

    The equation x2 = 1 has two solutions: one of them is positive (x = 1) and one of them is negative (x = -1).

    If you wrote this:
    x = [itex]\sqrt{1}[/itex]
    you are listing only one number; namely, 1.

    Every positive, real number has two square roots - a positive square root, and a negative square root. The expression [itex]\sqrt{x}[/itex], where x >= 0, represents the principal (or positive) square root of the number inside the radical.
     
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