1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Difference of two squares under a radical

  1. Sep 11, 2011 #1
    This may seem elementary, but it's been awhile since I've done heavy algebra work.

    I used the quadratic formula to get this equation:

    [itex]\mu = (\pm \sqrt{1 - 8x^{2}} - 1)/2[/itex]

    and I'm trying to simplify to:

    [itex]\mu = (\sqrt{1 + 8x^{2}} - 1)/2[/itex]

    If anyone could just point me to any resources that could give me the answer, or let me know of some obscure algebraic property I've forgotten, that'd be great!

    I'm pretty sure you can't distribute negatives into a square root, but x | x > 1.00, so any value of x would give a negative answer under the radical, and I'm left with an imaginary number.

    I also tried breaking down the difference of two squares into its factors, but that only works when I add another factor to compensate for 8, and makes it more messy than before.


    Thanks.
     
    Last edited: Sep 11, 2011
  2. jcsd
  3. Sep 11, 2011 #2
    Assuming the [itex]x[/itex] in both expressions is the same, then the two are equal if and only if:
    [tex]
    \mu = \frac{\pm \sqrt{1 - 8 x^{2}} - 1}{2} = \frac{\sqrt{1 + 8 x^{2}} - 1}{2}
    [/tex]

    [tex]
    \pm \sqrt{1 - 8 x^{2}} - 1 = \sqrt{1 + 8 x^{2}} - 1
    [/tex]

    [tex]
    \pm \sqrt{1 - 8 x^{2}} = \sqrt{1 + 8 x^{2}}
    [/tex]

    Up to this point all the transformations were equivalent. Now, we want to get rid of the root and we square the two sides. This may generate more solutions than the original equation had and we have to verify if each solution that we got satisfies the original equation:
    [tex]
    1 - 8 x^{2} = 1 + 8 x^{2}
    [/tex]

    [tex]
    16 x^{2} = 0
    [/tex]

    [tex]
    x^{2} = 0 \Leftrightarrow x = 0
    [/tex]
    This solution satisfies the original equation only in the case where the l.h.s. has the + sign!

    Notice that we only get a particular value for [itex]x[/itex] for which we can equate the two expressions. Thus, it is not an equality. I would say you either made an algebraic mistake in getting your original solution or the "new x" is related to the "old x" via (let's call the new x to be y):
    [tex]
    \mu = \frac{\pm \sqrt{1 - 8 x^{2}} - 1}{2} = \frac{\sqrt{1 + 8 y^{2}} - 1}{2}
    [/tex]

    [tex]
    \pm \sqrt{1 - 8 x^{2}} - 1 = \sqrt{1 + 8 y^{2}} - 1
    [/tex]

    [tex]
    \pm \sqrt{1 - 8 x^{2}} = \sqrt{1 + 8 y^{2}}
    [/tex]

    Squaring:
    [tex]
    1 - 8 x^{2} = 1 + 8 y^{2}
    [/tex]

    [tex]
    x^{2} + y^{2} = 0
    [/tex]

    In the set of real numbers, this relation has only a trivial solution [itex](x, y) = (0, 0)[/itex]. In the set of complex number it implies:
    [tex]
    y = \pm i \, x
    [/tex]
    and the equality holds only in the case of the "+" sign.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Difference of two squares under a radical
Loading...