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Let $x^2=x+1$, the solution is the golden ratio phi :$x=\frac{1\pm\sqrt{5}}{2}$.

Now I tried $x=x^2-1$ substituted in the equation itself : $$(x^2-1)^2=x^2\Rightarrow x^4-3x^2+1=0$$ $$\Rightarrow x=\pm\sqrt{\frac{3\pm\sqrt{5}}{2}}$$

Now we get 4 roots because one step is not iff giving back $x^2-1=\pm x$.

However we can recover the numerical values of the first equation solutions.

What happens now if I use $(x^2-1)^2=x+1\Rightarrow x(x^3-2x-1)=0$ ?

Could the numerical values be recovered in this case ? I do not think, so what step is used that is not reversible, only $x^2=x+1$ was used ?

QuantumPotatoïd
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  • you have to rule out the two cases were $x = -(x^2 -1)$. You added those to roots extraneeously. And $(\frac {1\pm \sqrt 5}2)^2 = \frac {6\pm 2\sqrt 5}4= \frac {3\pm \sqrt 5}2$ so... everything checks. $x =\sqrt {\frac {3\pm \sqrt 5}2}}$ and the solutions $x =\sqrt {\frac {3\pm \sqrt 5}2}}$ were the two extraneous solutions to $x = -(x^2 -1)$. – fleablood Jan 28 '21 at 20:53
  • "so what step is used that is not reversible" Simple. Squaring. Squaring is not reversable. $a=b \implies a^2 = b^2 \implies a = |b|\implies a = \pm b$. – fleablood Jan 29 '21 at 07:04
  • If the domain and codomain were set such that the function were uniquely reversible ? – QuantumPotatoïd Apr 11 '21 at 07:20

1 Answers1

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When you square $M = W$ to $M^2 = W^2$ you add in the extraneous solutions of $M = -W$.

So $x = (x^2 -1)$ has two solutions. And $x^2 =(x^2 -1)^2$ will have $4$ solutions: the two solutions to $x=x^2-1$ as well as two new and extraneous solutions to $x =-(x^2-1)$.

$x = \sqrt{\frac {3 \pm \sqrt 5}2} = \frac {1 \pm \sqrt 5}2$ are the two solutions to $x = x^2-1$.

$x = -\sqrt{\frac {3 \pm \sqrt 5}2} = -\frac {1 \pm \sqrt 5}2$ are the two solutions to $x = -(x^2 -1)$.

Notice that $( \frac {1 \pm \sqrt 5}2)^2=\frac {3 \pm \sqrt 5}2$ so $ \sqrt{\frac {3 \pm \sqrt 5}2} = \frac {1 \pm \sqrt 5}2$

fleablood
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