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I'll give an example for my first question:

$x^2 + x + 1 = 0$

Clearly $x = 0$ and $x = 1$ aren't solutions, so first we can safely divide by $x$:

$x + 1 + 1/x = 0$

By subtracting $1/x$ from both sides we get:

$x + 1 = -1/x$

By plugging the value $x + 1$ back we get:

$x^2 - 1/x = 0$

Multiplying by $x$ and adding $1$ to both sides:

$x^3 = 1$

Which $x = 1$ is clearly a solution to, unlike the original equation.

I have a problem with this, all manipulations did not include dividing by zero or any non-defined operations, all what was done is expressing $x$ by a different way so why does it change the final solution?

When we divided the equation by $x$ the solution shouldn't change which means the first equation holds so why isn't the second one compatible with the first?

On another note, I have a different question that is slightly related to this one, an example for this one is:

We have the system of equations $S$ which is:

$x + y + z = 1 \quad (L1)$

$x + y - z = 1/2 \quad (L2)$

$x - y + z = -4 \quad (L3)$

Then we transform this system into $S'$ by manipulating equations together:

$2x + 2y = 3 \quad (L1 + L2)$

$2y - 2z = 6 \quad (L2 - L3)$

$2x + 2z = -3 \quad (L1 + L3)$

The solution for $S$ (which is $(-3/2, 5/2, 0)$) also doesn't satisfy $S'$, which is counterintuitive to me because that's the kind of transformations we're taught then why does it change the solution?

For both questions are those like general phenomenons or are they just special cases? What are the names of topics concerned with the act of transforming equations like the examples above?

Abdullah Alhussni
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  • What do you mean by “plugging the value $x+1$ back”? – José Carlos Santos Aug 10 '20 at 18:01
  • @JoséCarlosSantos: Substituting $-\frac1x$ for $x+1$ in the original quadratic. – Brian M. Scott Aug 10 '20 at 18:01
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    It's like you multiplied by $(x-1)$ implicitly. $x^3-1=(x-1)(x^2+x+1)$ – Alexey Burdin Aug 10 '20 at 18:03
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    For the "plugging thing", you are essentially doing a subtraction of two equations. https://math.stackexchange.com/questions/1471713/imposter-solution-for-sqrt2x15-6-x This might be helpful. – cr001 Aug 10 '20 at 18:03
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    $x=1$ is a solution to your new equation but it's irrelevant because we are specifically looking at the values of $x$ which are not $1$ or $0$. If you allow $x=1$ then it's like you multiplied both sides of your equation by $0$. – Jürgen Sukumaran Aug 10 '20 at 18:07
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    $1=0$ has no solutions but multiplying by $x$ gives $x=0$ which has one solution. In the second question, $(L2-L3)$ should be $2y-2z=9/2$ (also $(L1-L2)$ is wrong). – pancini Aug 10 '20 at 18:08
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    For she system of equations you can add or subtract equations, but step by step. You can't modify all equations simultaneously like $Ax=b\Rightarrow BAx=Bb$ unless $\det B\ne 0$. In your case $B=\begin{pmatrix}1&1&0\0&1&-1\1&0&1\end{pmatrix}$ and $\det B=0$. – Alexey Burdin Aug 10 '20 at 18:13
  • Thanks guys! Your comments really helped. – Abdullah Alhussni Aug 10 '20 at 18:14
  • Actually I copied it from my school textbook, I didn't double check but this is exactly how it's written, but I'm not sure where the problem in the book is but nonetheless they didn't explain what the issue was. @ElliotG – Abdullah Alhussni Aug 10 '20 at 18:16
  • why don't L1+L2 and L2-L3 have fractions at the right hand side? – user13267 Aug 11 '20 at 02:55
  • @user13267 Both sides are multiplied by 2 in all equations I think (or at least they're simplified) – Abdullah Alhussni Aug 11 '20 at 09:34
  • they're mistakes – user13267 Aug 11 '20 at 10:58
  • @user13267 okay if we fixed the system $S'$ does it change the fact that $S$'s solution does not satisfy $S'$? – Abdullah Alhussni Aug 11 '20 at 16:53

3 Answers3

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The problem is that $x+1$ is not in general equal to $-\frac1x$: that equality holds specifically for solutions to the original equation $x^2+x+1=0$. Thus, the new equation is not in general equivalent to the original quadratic: they are equal only when $x$ is already a solution to the original quadratic. Since $x=1$ is not such a solution, the fact that it is a solution to the new equation is irrelevant to the original problem.

In effect you multiplied by $x-1$ when you converted the original quadratic to a cubic, thereby introducing the extraneous solution $x=1$: $$(x-1)(x^2+x+1)=x^3-1\,,$$ so $(x-1)(x^2+x+1)=0$ iff $x^3-1=0$.

Brian M. Scott
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3

You may not substitute parts of an equation into itself. E.g. if $x^2=x$, the substitution of $x^2$ by $x$, gives $x=x$ !

  • What's the explanation for this? Also what about substituting $x^2$ into $x^2$ as in ($x^4 = x$)? – Abdullah Alhussni Aug 10 '20 at 18:49
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    @AbdullahAlHussni: after such manipulations, the solutions of the initial equation remain perforce solutions, but by transforming the expressions in a non-reversible way you can introduce alien solutions. Just like $x=1\implies 0\cdot x=0\cdot 1$ is non reversible. –  Aug 10 '20 at 19:03
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$$\begin{aligned} a\left( x \right) &= x^2 + x + 1\\ \frac{1}{x}a(x)&=x+1+\frac{1}{x}\\ a(x)-\frac{1}{x}a(x) &=\color{red}{(1-\frac{1}{x})}a(x)=x^2-\frac{1}{x}\\ \therefore \color{red}{(1-\frac{1}{x})}&=0\text{ is the additional root} \end{aligned} $$

Forsteri123
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