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Solve y=x^6 for x

  1. Oct 24, 2008 #1
    1) We all know that y=x2 => x=+/- sqrt(y) and z2=9=> z=+/-3.

    But does this idea extend to all even exponents?

    e.g. Assume the field of real numbers, solve y=x6 for x.
    Is the answer x=+/-y1/6? Is this true in general for any even exponent? (e.g. y=x28 => x=+/-y1/28 ?)

    2) x2 = 9
    (x2)1/2 = 91/2 (apply the SAME operations on BOTH sides)
    x1 = x = 3 (by applying the exponent law: power of a power (xa)b=xab )
    This is certainly wrong (the correct answer shold be +/- 3), but I don't see where the mistake is. Can someone please help me?

    Thanks for explaining!
    Last edited: Oct 24, 2008
  2. jcsd
  3. Oct 24, 2008 #2
    Yes it does.

    You are making a mistake a lot of people make with square roots. Namely, you are acting as if the square root operation "undoes" squaring. That [tex]\sqrt{x^2} = x[/tex] for all x. This isn't a true statement, though! Take x = -1. [tex]\sqrt{(-1)^2} = \sqrt{1} = 1 \neq -1[/tex].

    Of course, square root DOES undo squaring *sometimes*. For positive numbers and for zero, it IS true that [tex]\sqrt{x^2} = x[/tex].

    Note also that [tex]\pm[/tex] is just a shorthand notation. It is an abbreviation for something that CANNOT be expressed in a single equation. When we say "ecks equals plus or minus 3" we don't mean it's equal to a value called [tex]\pm 3[/tex]. And x is a single variable, and can only take on one value at a time. It's technically more correct to say "the values of x for which [tex]x^2 = 9[/tex] are +3 and -3. Or, using logic notation we notate this as [tex]x = 3\vee x = -3[/tex]" (the [tex]\vee[/tex] character means "or"). We don't know which value x really is, but we know it is one of the two. If the problem continues past this point, we have to take each solution into account on a case-by-case basis.

    So take home messages:
    * Square root only "undoes" squaring for non-negative numbers.
    * Square roots are always positive.
    * When looking for solutions to an equation, if there is more than one solution, we cannot properly write it in a single equation.
    * "[tex]x = \pm y[/tex]" is just a shorthand for "[tex]x = -y[/tex] or [tex]x = y[/tex]".
  4. Oct 24, 2008 #3


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    Very nicely said!

    (You left me nothing to say, but I just have to put my oar in!)
  5. Oct 25, 2008 #4
    Thanks! This is very helpful comments to keep in mind!

    More questions:

    Consider the steps:
    x^2 = 9
    (x^2)^1/2 = 9^1/2 (apply the SAME operations on BOTH sides)
    x = 3 (by applying the exponent law: power of a power (x^a)^b=x^(ab) )

    Fact 1: I was told that in equations, if you apply the SAME operations to BOTH sides (e.g. log both sides, +2 to both sides, x2 to both sides, etc.), the equality still holds. [this is what I am doing on the 2nd line]

    Fact 2: exponent law: power of a power [I simply applied this law on the 3rd line]

    But I got the wrong answer, so does this mean either fact 1 or fact 2 is wrong? If so, which one is wrong?
  6. Oct 25, 2008 #5


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    They are both perfectly correct. But that only tells you that x= 3 is one possible solution of x2= 9.
  7. Oct 26, 2008 #6
    Your logic is well reasoned at all steps, and the result is a true statement.

    But your teacher wasn't asking for your answer to be true. The question was probably phrased something like "find all possible values for x such that x^2 = 9". While you arrived at one possible value, you did not provide all the possible values.

    Problems of the form "find all values that satisfy this equation" appear all the time in math. It's not hard to prove that in this case there are only two solutions. As you come across different problems of this kind, you'll learn more techniques to make sure that you have found all the solutions.
  8. Oct 26, 2008 #7
    So in other words, the find ALL solutions we need to have "if and only if" at every step, right?
    i.e. x^2=9

    If the above is true, then I am pretty sure that in the following steps, the implication only goes in 1 direction, which direction is it? (<= or =>?)
    x^2 = 9
    (x^2)^1/2 = 9^1/2
    x = 3

    x^2 = 9
    => (x^2)^1/2 = 9^1/2
    => x = 3 ??? or the other way around?

    Thanks for explaining!
  9. Oct 26, 2008 #8


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    If x= 3 then x2= 9

    If x= -3 then x2= 9

    are both true. There for "if x2= 9 then x= 3" is untrue.
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