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I Is there any solution to the equation x^(0.5)+1=0?

  1. Apr 19, 2017 #1
    I got myselft wondering if there is any solution to the equation x^(0.5)+1=0. I know that for a real x there is not but, when I assume x is any imaginary number in the form e^(ix/2) and then solve the equation e^(ix/2)=-1 the result is x=2(2 pi n + pi) for integer values of n. If one then takes n=1, x=6pi and e^(i 6pi) = 1. Whaaaat?

    My other question is why in an equation like the one above, squaring both sides to get the solution is not allowed since it would result in x=1.
     
  2. jcsd
  3. Apr 19, 2017 #2
    Why isn't ##1## a solution ?
     
  4. Apr 19, 2017 #3

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hello Alan, :welcome:

    You mean 'of the form ##e^{ix}## '
    You mean 'and ##e^{3\pi} = -1## '

    Question is: is ##e^{6\pi}## in the domain of the function, or is the domain limited to arguments ##[0, 2\pi)## or perhaps ##(-\pi , \pi]## ? Read about it here

    (I only get to see the animation when I open the picture in a new tab)
     
  5. Apr 19, 2017 #4
    Well I think that once you substitute x=1 in the equation you would get:

    1^(0.5) + 1 = 0
    2=0 ????

    Am I missing something?
     
  6. Apr 19, 2017 #5
    Thank you for the answer and the greeting! Let me put it differently:

    I have the equation:

    ##x^{0.5}## +1 = 0

    So I assume that ##x=e^{iy}## and then I get that ##e^{iy/2}+1=0##

    From the Euler's identity I have a solution in the form ##y=2\pi##

    So ##x=e^{2\pi i}## which gives ##x=1## and therefore ##1^{0.5}+1=0##. Where am I missing something?
     
  7. Apr 19, 2017 #6
    No you are correct but also ##(-1)^2 = 1## so ##1^{1/2} = \pm 1##. Or ##1^{1/2} + 1 = -1 + 1 = 0##.

    You were told to find the solutions to the equation.
     
  8. Apr 19, 2017 #7
    But if you graph the function ##y(x)=1^{x}## and look at the point ##x=0.5## you will get y=1. In the other hand, if you graph ##(-1)^{x}## you will get ##y=i##. So how can you affirm that ##1^{1/2}+1=-1+1=0##??
     
  9. Apr 19, 2017 #8
    Are you saying that you don't believe that ##(-1)^2 = 1## ?

    When you graph a function, the graph will be on the domain and range of the function. For the function ##f(x) = \sqrt{x}##, the domain and range is ##x\ge 0## (by convention for range).

    You are given to find the solutions to a equation not roots of a function ##f(x) = \sqrt{x} + 1##.
     
  10. Apr 19, 2017 #9

    Mark44

    Staff: Mentor

    The graph of ##y = 1^x## isn't very interesting, assuming x and y are real numbers. 1 raised to any finite power is just 1, so the graph is just a horizontal line through (0, 1).
     
    Last edited: Apr 19, 2017
  11. Apr 19, 2017 #10

    Mark44

    Staff: Mentor

    By convention ##\sqrt{1} = + 1##, ##\sqrt 4 = + 2##, NOT ##\pm 1## or ##\pm 2##, respectively. By extension, ##1^{1/2} = \sqrt 1##.
     
  12. Apr 19, 2017 #11

    Mark44

    Staff: Mentor

    If ##z = e^{i\cdot 2\pi}## = 1 + 0i, then the principal (complex) square root is ##e^{i \cdot \pi}## = -1 + 0i. So this value of z is a solution to the equation ##z^{1/2} = -1##, or equivalently ##z^{1/2} + 1 = 0##.

    I am using z to reinforce the idea that I'm working with complex numbers.
     
  13. Apr 19, 2017 #12
    When you wirte ##z=-1+0i## aren't you saying this is a pure real number once the imaginary part is zero? If so, is it right to say that the equation has a real solution. This is very confusing to me.
     
  14. Apr 19, 2017 #13

    Mark44

    Staff: Mentor

    z here has no imaginary part, but ##z^{1/2}## is complex, and that is what the equation involves.
     
  15. Apr 19, 2017 #14
    Reason for that convention is that to fit ##\sqrt{x}## in functions but that does not matter when solving a equation.
     
  16. Apr 19, 2017 #15

    Mark44

    Staff: Mentor

    You are mistaken.
    In the real numbers, ##\sqrt x = 2## has one solution -- x = 4.
     
  17. Apr 19, 2017 #16

    pwsnafu

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    Science Advisor

    No, it comes from solving equations.

    Suppose we are given ##a \geq 0## and we wish to find ##x \in \mathbb{R}## such that ##x^2 = a##. Then we show:
    1. If ##a = 0## there only exists one solution, namely ##x=0##,
    2. If ##a \neq 0## then there exists two distinct solutions ##x_1## and ##x_2##,
    3. We involke the fact that the reals have a total ordering, so that either ##x_1 > x_2## or ##x_1 < x_2##.
    4. We prove that ##x_2 = -x_1##.
    After doing all that, we conclude that one of the two solutions is positive and the other is negative. We define the notation ##\sqrt{a}## to be the positive value and call it the principle square root. Due to 4, the other root is then ##-\sqrt{a}##.

    Further, it's easy to see that "that does not matter when solving a equation" is a false statement by just looking at the quadratic formula. If ##\sqrt{a}## meant both positive and negative the formula would never have the ##\pm## sign in it.
     
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