# Simple equation causing a lot of headache!

1. Jan 23, 2009

### Marin

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:

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

assume y=z=0:

Now, the equation becomes:

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

I guessed a solution at x=0

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

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

Then the equation becomes:

(*) $$(x-a)(x-a)^{-3}+(x+a)(x+a)^{-3}=0$$, or

$$(x-a)^{-2}+(x+a)^{-2}=0$$

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

$$(-a)^{-2}+(+a)^{-2}=1/a^2+1/a^2$$ 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: $$(-a)^{-4}+(+a)^{-2}=1/a^4+1/a^2$$ 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!

2. Jan 23, 2009

### shirel

I didn't read your post right through, but still I encountered 2 difficulties:
1) How could you assume that y=z=0?
2) In order to solve equation, which includes 3 variables, you need another 2 equations..

3. Jan 23, 2009

### Staff: Mentor

Looks like a sum of fractions to me. Why don't you add them first?

4. Jan 23, 2009

### wsalem

So what! This shows that your assumption "x=0 is a solution" was incorrect. Also (x-a) doesn't have to be zero either.

Assuming y,z = 0. Starting from the equation
$$\frac{1}{(x-a)^{2}} + \frac{1}{(x+a)^{2}}=0$$
$$=\frac{(x-a)^{2}+(x+a)^{2}}{(x-a)^{2}(x+a)^{2}} = \frac{2x^2 + 2a^2}{(x-a)^{2}(x+a)^{2}}$$
Multiply by $$(x-a)^{2}(x+a)^{2}$$ then dividing by two, finally
$$x^2 + a^2 = 0$$
which have a complex solution $$x = \sqrt(-a^2)$$

Last edited: Jan 23, 2009
5. Jan 23, 2009

### Staff: Mentor

Nah, I thought it cancels out, but I was on the wrong track. Brain slip.

6. Jan 23, 2009

### Marin

ok, I spared you some details, it´s a part of a system of equations but the other two just give y=z=0, in fact it´s the fist derivative in X to the potential of two point charges q in the distance a from the origin.

I cannot add fractions here, because the denominators are different. And it´s kind a difficult to try to make one fraction out of the two, cuz there are these roots of cubes there.. that´s why I tried guessing...

However x=0 proves to be a solution if you don´t try to simplify. That´s why I suspect that this solution gets lost after the simplification.. but how is it possible?

7. Jan 23, 2009

### Staff: Mentor

Find common denominator, once you get one fraction it can be zero only when nominator is zero. But it doesn't look like this nominator is easier to solve.

8. Jan 23, 2009

### flatmaster

I see an error in the first solution. In the step affter you assumed x=0, You distributed an a over parenthesis and took thoes terms to the other side of the eaquation in one step. You missed a negative sign for the second term.

9. Jan 23, 2009

### Marin

If you mean the equations after "check" I olny use that (-a)²=a². I don´t do anything with the second term, just rewrote it.

10. Jan 23, 2009

### wsalem

Can you elaborate on that please! How does it prove to be a solution? and why $$\sqrt(-a^2)$$ isn't a solution!

11. Jan 24, 2009

### Marin

Well, I mean, when you plug x=0 in the equation it works until you simplify too much, and contract factors, I think it lies ot this square (-a)² under the root, cuz when you contract it you leave the - there, and x=0 is no more a solution, but if you first get rid of the minus, then contract, x=0 will be a solution.

sqrt(-a²) is a possible solution, but to be honest, I don´t like it, because the potential which the function stems from is defined in R³, not in C³ :(

12. Jan 24, 2009

### wsalem

It is not like what we want is what we will get! You see just because the equation is defined on R doesn't mean its solution must exist there? R is not algebraically closed.

No, you said
Prove it! Show how would x=0 reduced the equation to zero?
Edit: Sorry, been looking over $$\frac{1}{(x-a)^{2}} + \frac{1}{(x+a)^{2}}=0$$ without realizing the mistake in the previous derivation. 0 is indeed a solution as pointed out by gabbagabbahey

Last edited: Jan 24, 2009
13. Jan 24, 2009

### gabbagabbahey

$x=0$ is a perfectly valid solution. Marin correctly showed it was a solution in his original post using his first method.

Hi Marin, your error is in this step:

Remember what the fractional exponent means: $u^{3/2}=\sqrt{u^3}$ which is always positive. When you take the square root of $(x-a)^2$, you only get $x-a$ if that quantity is positive. For example, suppose you had $x=2$ and $a=3$ , then you would find

$$\left((x-a)^2\right)^{-3/2}=\left((2-3)^2\right)^{-3/2}=\left((-1)^2\right)^{-3/2}=\left(1\right)^{-3/2}=1$$

But if you blindly apply the exponent rule without taking care to take the positive root, you get:

$$\left((x-a)^2\right)^{-3/2}=(x-a)^{-3}=(-1)^{-3}=-1$$

The easiest way to account for this is to use absolute value signs:

$$\left((x-a)^2\right)^{-3/2}=|(x-a)|^{-3}$$

Last edited: Jan 24, 2009
14. Jan 24, 2009

### Marin

Great!

I knew it was there!

Thanks a lot, all of you! Now it´s all clear :)