Simple equation causing a lot of head ache!

  • Thread starter Marin
  • Start date
  • #1
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Hi there!

I was trying to solve an equation but got very perplexed by the fact that a certain number x=0 is both a solution and no solution:

Here´s the equation:

[tex](x-a)[(x-a)^2+y^2+z^2]^{-3/2}+(x+a)[(x+a)^2+y^2+z^2]^{-3/2}=0[/tex]

assume y=z=0:

Now, the equation becomes:

[tex](x-a)[(x-a)^2]^{-3/2}+(x+a)[(x+a)^2]^{-3/2}=0[/tex]

I guessed a solution at x=0

check: [tex]-a[(-a)^2]^{-3/2}+a[a^2]^{-3/2}=0[/tex], ok I assume it´s true

Now let´s use the exponent rule: (a^x)^y=a^(xy)

Then the equation becomes:

(*) [tex](x-a)(x-a)^{-3}+(x+a)(x+a)^{-3}=0[/tex], or

[tex](x-a)^{-2}+(x+a)^{-2}=0[/tex]

Ok, plug once again x=0 and there coems the surprise:

[tex](-a)^{-2}+(+a)^{-2}=1/a^2+1/a^2[/tex] is not equal to 0!!!

What is more, if you substitute x=0 before you contract the factors in (*) ypu come up with: [tex](-a)^{-4}+(+a)^{-2}=1/a^4+1/a^2[/tex] again no 0, but also different from the result above!


I suppose I´m doing it somehow wrong woth the powers and exponents but I can´t figure it out!

If you see my mistake, please tell me!

Thanks in advance!
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi Marin! :smile:
… Now let´s use the exponent rule: (a^x)^y=a^(xy)
Sorry … doesn't always work, especially if y is a fraction and a is negative :wink:
 

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