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I have tried to solve this question, and I found the one real root, x=0, but I do not understand how to show that there is only one root: Find the number of real solutions of $x+\sqrt{x^2+\sqrt{x^3+1}}=1$
The solution to the question, is 1 real solution, but I am not sure how to prove that there are no other solutions.

frog0101
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    You solve the equation by squaring twice, remembering that this may introduce extraneous solutions. You will find two real solutions, but one will be extraneous, leaving only one solution. – John Wayland Bales Mar 25 '18 at 05:31

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For the real square root to be defined $\,\sqrt{x^3+1}\,$, it is necessary that $\,x^3+1 \ge 0 \iff x \ge -1\,$. For $\,\sqrt{x^2+\sqrt{x^3+1}}=1-x \ge 0\,$, it is necessary that $\,x \le 1\,$, so any real solution is in $\,[-1,1]\,$.

Eliminating the radicals:

$$ x^3+1 = \left((1-x)^2-x^2\right)^2 \;\;\iff\;\; x^3 - 4 x^2 + 4 x = 0 \;\;\iff\;\; x(x-2)^2=0 $$

The latter has the unique real solution $\,x=0\,$ in $\,[-1,1]\,$.

dxiv
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