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Solving the following equation for x: ##(x-5)^{-2} = 9##

  1. Jun 24, 2016 #1
    1. The problem statement, all variables and given/known data
    An exercise in Lang's Basic Mathematics which asked to solve for x in the equation ##(x - 5)^{-2} = 9##

    I tried to solve it and got ##x = \frac {16} 3##

    which is correct according to the book, but I've found that the book states that x also equals \frac {14}3, and I can't figure out how.
    2. Relevant equations


    3. The attempt at a solution
    ##(x-5)^{-2} = 9##

    This yields: ##x-5 = 9^{-1/2} =\frac 1 {9^{1/2}} = \frac13##

    Which yields: ##x=\frac 13 +5 = \frac {1+15} 3 =\frac {16}3##

    But I can't figure out how can x equals ##\frac {14}3## too.
     
  2. jcsd
  3. Jun 24, 2016 #2

    Mark44

    Staff: Mentor

    It's better to write this as ##(x - 5)^2 = \frac 1 9##, which leads to ##x - 5 = \pm \sqrt{\frac 1 9}##
    Quadratic equations can have two solutions.
     
  4. Jun 24, 2016 #3
    What is ##(-\frac{1}{3})^2##?
     
  5. Jun 24, 2016 #4
    I am thinking about that old math movie with young actor Edward James Olmos. He was playing the role of high school teacher.

    Stand and deliver

    "Negative times negative is positive!!!" The result would be 1/9


    For the original poster's benefit. The following is compiled from my own high school era old math factbook.

    second degree equations.
    • you can evaluate whether the quadratic has two roots, or one root, or none of the real-numbered roots. (roots belonging to ℝ)
    • calculate the quadratic into the form of: $$Ax^2 + Bx+C= 0$$
    $$x=\frac{-B±\sqrt{B^2-4AC}}{2A}$$

    this is called the discriminant part of the equation ##D=\sqrt{B^2-4AC}##

    if D>0, then the equation has two non-equal roots
    if D=0, then the equation has a twin-root
    if D<0, then the equation doesn't have real-numbered roots (roots in the real numbers ℝ)
    This last fact is because negative inside the squareroot sign is not defined.
    E.g. ##\sqrt{(-5)}= ~~not ~~defined~~ in~~ real~~ numbers##

    negative inside the squareroot is defined in the complex numbers, I think, but I have no knowledge of the complex numbers myself. (complex numbers symbol is this one ℂ)
     
  6. Jun 24, 2016 #5

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    In short if A2 = B, then A = ± √B
     
  7. Jun 24, 2016 #6
    I got it now, thanks everyone. As usual the solution is simpler than expected. My mistake was not considering ##-3 = 9^{1/2}##
     
  8. Jun 24, 2016 #7

    Mark44

    Staff: Mentor

    The symbol ##\sqrt{x}## denotes the positive square root of x, by established convention. In this case ##\sqrt{9} = +3##. However, the equation ##x^2 = 9## has two solutions: 3 and -3.
     
  9. Jun 24, 2016 #8

    Ray Vickson

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    Science Advisor
    Homework Helper

    No: ##9^{1/2}## is ##+3##, NOT ##-3##. Essentially by definition, the functions ##x^{1/2}## and ##\sqrt{x}## mean the positive root.

    There is no really good language for it, but saying "a" square root instead of "the" square root comes close to capturing it: there are two roots to the equation ##x^2 = a \:(a > 0)##; these are ##x = \sqrt{a}## and ##x = -\sqrt{a}##
     
  10. Jun 24, 2016 #9

    • watch this khan academy video it will be helpful for this problem and also for the purpose of increased understanding.
    • review what is the binomial square formula such as:
    • ##(a+b)^2= a^2+2ab+b^2##
    • ##(a-b)^2 = a^2-2ab+b^2##
     
  11. Jul 6, 2016 #10

    James R

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    Alternatively you could do this:
    ##\frac{1}{(x-5)^2}=9##
    ##9(x-5)^2 = 1##
    ##9(x^2 - 10 x + 25) = 1##
    ##9x^2 - 90 x + 224 = 0##
    ##(3x - 14)(3x - 16)=0##
    ##x=\frac{14}{3}## or ##x=\frac{16}{3}##.
     
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