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Question: $$\frac{x}{\sqrt{x^2+1}} = x^4 - x$$

I tried: $$\rightarrow \frac{1}{\sqrt{x^2+1}} = x^ 3 - 1$$

$$\to\frac{\sqrt{x^2 + 1}+1}{\sqrt{x^2+1}} = x^3$$

Now rationalising it $$\to \frac{x^2 +1-1}{x^2+1-\sqrt{x^2+1}} = x^3$$

$$\frac{1}{x^2+1-\sqrt{x^2+1}} = x$$

Can we do anything with this? How to solve? Please help me.

  • $\rightarrow \frac{1}{\sqrt{x^2+1}} = x^ 3 - 1$,After this square on both sides and then cross multiply,you will get degree 6 polynomial.. –  Jan 25 '22 at 10:47
  • Yeah that 's true.,but that seems more solvable than the current state i guess. –  Jan 25 '22 at 10:48
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    from the first line to the second, you can't just simplify by $x$. What if $x = 0$! – Essaidi Jan 25 '22 at 10:51
  • Not sure it helps a lot... but by monotonicity arguments, you can notice that the equation has a unique non-zero real solution. – mathcounterexamples.net Jan 25 '22 at 10:51
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    Yes, if we simplify by $x$ we lose the fact that $0$ is a obvious solution. – Essaidi Jan 25 '22 at 10:56
  • @Anonymous Where is the problem coming from? – mathcounterexamples.net Jan 25 '22 at 10:59
  • @Anonymous: you use the various comments: $x=0$ is an obvious solution, and the other is between $1$ and $2$ and the root of a degree $6$ polynomial (you need to be slightly careful, as squaring introduces a spurious root between $0$ and $1$). You can find a value for the second solution using numerical methods – Henry Jan 25 '22 at 11:02

1 Answers1

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From the graph of the function : $$f(x) = \dfrac{1}{x^2 + 1 - \sqrt{x^2 + 1}} - x$$ we see that there is only one other solution $x = 1.181$.
We deduce that solutions of the equation are $0$ and $1.181$.

Essaidi
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  • It's not always possible to solve equations by elementary tools. In your case solving polynomial of high degree is impossible in general, that's why we use numerical approach. Like plotting the graph. – Essaidi Jan 25 '22 at 11:09
  • As was mentioned in the comments of the question, the only solutions are not this one. First you need to consider $x=0$ when you divide by $x$. But $0$ is one solution. Then, for simplicity, just square both side of the equation instead of doing all your work here. You will need to check after your calculus which solutions fits the initial one, but it is simpler in my opinion (as stated again in the comments of the question). – Martigan Jan 25 '22 at 12:52