- #1

Marin

- 193

- 0

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}=-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 comes the surprise:

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

What is more, if you make the substitution x=0 in (*), i.e. before you contract terms, you end up with: [tex](-a)^{-4}+(+a)^{-2}=1/a^4+1/a^2[/tex] again not equal to 0, but an entirely different expression!

I guess I´m doing it wrong with the powers and exponents, but I cannot figure out where my mistakes are.

If you see it, please be kind and tell me!

Thanks a lot!