# Simple equation causing a lot of headache

• Marin
In summary, the conversation discusses an equation that has a solution of x=0 but becomes inconsistent after simplification. The participants debate the validity of this solution and potential errors made in solving the equation. The main point of contention is an error in using the exponent rule, which can be resolved by accounting for the positive root of a fractional exponent using absolute value.
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!

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..

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

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

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

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, because 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?

Marin said:
nd it´s kind a difficult to try to make one fraction out of the two, because there are these roots of cubes there.

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.

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.

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.

However x=0 proves to be a solution if you don´t try to simplify.
Can you elaborate on that please! How does it prove to be a solution? and why $$\sqrt(-a^2)$$ isn't a solution!

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, because 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 :(

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 :(
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
However x=0 proves to be a solution

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:
wsalem said:
So what! This shows that your assumption "x=0 is a solution" was incorrect.
$x=0$ is a perfectly valid solution. Marin correctly showed it was a solution in his original post using his first method.

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?

Hi Marin, your error is in this step:

Marin said:
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

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

!
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:
Great!

I knew it was there!

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

## 1. What is a simple equation causing a lot of headache?

The "simple equation causing a lot of headache" refers to a mathematical equation that appears to be straightforward and easy to solve, but actually leads to unexpected and complicated results.

## 2. Why is this equation causing so much trouble?

This equation is causing trouble because it often involves multiple variables and unknowns, making it difficult to find a definitive solution. Additionally, small errors in calculation or assumptions can greatly affect the final result.

## 3. Can you give an example of this type of equation?

One example of a simple equation causing a lot of headache is the quadratic equation, which is used to solve for the roots of a quadratic polynomial. While the equation itself is simple, the process of finding the roots can become quite complex and lead to unexpected solutions.

## 4. How do scientists deal with these types of equations?

Scientists use various mathematical techniques, such as factoring, substitution, and graphing, to solve these types of equations. They also carefully check their work and make sure to consider all possible solutions.

## 5. Is there a way to avoid these equations causing headaches?

While there is no guaranteed way to avoid these types of equations, scientists can use caution and double-check their work to minimize errors. They can also seek help from colleagues or use advanced mathematical software to assist with solving complex equations.

• General Math
Replies
9
Views
1K
• General Math
Replies
3
Views
1K
• General Math
Replies
8
Views
2K
• General Math
Replies
7
Views
995
• General Math
Replies
7
Views
1K
• General Math
Replies
2
Views
782
• General Math
Replies
2
Views
825
• General Math
Replies
4
Views
1K
• General Math
Replies
4
Views
1K
• General Math
Replies
11
Views
941