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WHat is this?

  1. Aug 20, 2006 #1
    ok say i have y^2 = 5x
    what does y=?
    Last edited: Aug 20, 2006
  2. jcsd
  3. Aug 20, 2006 #2
    I may be missing your question... but y = sqrt(5x).

    Unless I'm missing your point.
  4. Aug 20, 2006 #3
    thanks...soz.... im getting confused whether it was + or - sqrt(5x), or just sqrt(5x)
  5. Aug 20, 2006 #4
    You can't tell y is the positive or negative square root of 5x from the information given. For every value of x except zero, there are two distinct values of y that satisfy y^2 = 5x. They will always be additive inverses of each other as well.
  6. Sep 12, 2006 #5
    When you balance out a square number you always put a +/- with the square root. This is because using this example:
    x^2 = 1 => x = +-1
    both +1 and -1 squared gives 1, so x can be both +1 or -1, hence the x = +-1
  7. Sep 13, 2006 #6
    it's [tex]\sqrt{5x}[/tex] of course. The square root undoes the "Squared" but because you write the [tex]\sqrt{ ... }[/tex] yourself you must accept both positive and negative values.
  8. Sep 13, 2006 #7


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    I'm not sure what "accept both positive and negative values" means nor what the fact that you write it yourself has to do with it but:
    [itex]\sqrt{5x}[/itex] is by definition the positive number whose square is 5x. If you want to "accept" both positive and negative values you will need to write [itex]y= \pm \sqrt{5x}[/itex].
  9. Sep 13, 2006 #8
    What I mean (and my math teacher struggled to make this one go in) is that if you see a problem like [tex]y=\sqrt{x}[/tex] you just draw its positive. But if you see [tex]y^{2}=x[/tex] you take both positive and negative.

    My problem is with notation to be honest. Because I can take the square root of a number and end up negative...let's say [tex]\sqrt{25} = -5[/tex]. I mean it's probably incorrect due to some deffinitions which I don't know but if I check it, by working backwards, [tex] (-5)^{2} = 25[/tex]. And since math properties (which I cant name either) make it so doing something in left or exact opposite in right side to get you to same answer, well I don't see why I have to add a -in front of a sqrt because...the sqrt itself can spit out in my oppinion its own negative.

    Edit: However I've seen the raising to a power as somthing including logs or natural logs...and for that you'd need to have positives. I'm assuming that is why it's correct to choose to add a +/- ?
    Last edited: Sep 13, 2006
  10. Sep 14, 2006 #9
    No. The reason that we have to put a +/- sign in front of the square root when solving y2=5 is because we want to define square root to be a function, and a function cannot have more than one output for the same input. Thus, if you take Sqrt(25) you always get 5, never -5.

    But since the positive root of 5 is only one of the solutions to the equation, you should mark that the negative root of 5 is also a solution. Hence the +/- sign.

    EDIT: And since you mentioned the Log (that's the natural log) based definition of square root (indeed, all complex powers of complex numbers), you have to understand that this runs into the exact same problem as the real square root.

    The complex function Log(z) is defined to be the inverse operation of the complex exponential function, exp(z) (or ez). Unfortunately, this is not a good definition, because, like x2, exp(z) is not one-to-one. In fact, exp(z) is periodic with period [tex]2\pi i[/tex]. Thus, we have to restrict the domain of exp(z) in order to make Log(z) a function (i.e. having only one value per input). This results in the square root being positive rather than negative. But if we restrict the domain of exp(z) differently than the standard definition, our "log" function will result in different outputs for some or all values of z. Thus, you could conceivably have the square root be negative instead, or be negative for rationals and positive for irrationals (although that branch of the logarithm can't be pleasent).
    Last edited: Sep 14, 2006
  11. Sep 14, 2006 #10
    exactly. If [itex]a[/itex] is a positive real number, then it has two real square roots: one negative, and one positive. We define [itex]\sqrt{a}[/itex] to be the positive root. So [itex]y^2 = a[/itex] has two solutions, [itex]y = \sqrt{a}[/itex] (the positive root) and [itex]y = -\sqrt{a}[/itex] (the negative root), and you can succinctly write [itex]y = \pm \sqrt{a}.[/itex]
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