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Solve in $R$ the following equation :

$2(2(x^{2}-2)^{2}-7)^{2}-1=\sqrt{1-x^2}$

$x=0$ clearly one solution , but how I find other solutions How many solutions?

My attempt is put

$ y =\sqrt{1-x ^ 2} $

But I do not know how to solve this type of equations?

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    $\begingroup$ I agree squaring can introduce extraneous solutions, but I disagree that $0$ is the only solution; see the answer by Claude Leibovici $\endgroup$ – J. W. Tanner Jul 9 '19 at 10:32
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You must take care that squaring processes introduce extra roots.

In the present case, using the symmetry, I should simply plot the function $$f(x)=2(2(x^{2}-2)^{2}-7)^{2}-1-\sqrt{1-x^2}$$ for $0 \leq x \leq 1$ and visually notice that there is a root close to $x=0.5$.

To approximate the root, use a Taylor expansion around this point to get $$f(x)=\left(\frac{17}{32}-\frac{\sqrt{3}}{2}\right)+\left(\frac{49}{2}+\frac{1}{\sqrt{3}} \right) \left(x-\frac{1}{2}\right)+O\left(\left(x-\frac{1}{2}\right)^2\right)$$ Ignoring the higher order terms, this gives as an estimate $$x=\frac 12+\frac{3 \left(16 \sqrt{3}-17\right)}{16 \left(147+2 \sqrt{3}\right)}\approx 0.513350$$ while the solution is $\approx0.512603$.

Added for your curiosity

We could get a better approximation if, instead of using Taylor expansion, we use the $[1,1]$ Padé approximant. This would give as an estimate $$x=\frac 12+\frac{6954 \sqrt{3}-7209}{310977+42536 \sqrt{3}}\approx 0.512572$$

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    $\begingroup$ By the symmetry you mentioned, if $x$ is a root, then so is $-x$ $\endgroup$ – J. W. Tanner Jul 9 '19 at 10:28
  • $\begingroup$ @J.W.Tanner. For sure ! I supposed (may be wrongly) that it was explicit. $\endgroup$ – Claude Leibovici Jul 9 '19 at 10:38
  • $\begingroup$ Maybe it was; I'm not sure; anyway, yours is the only answer I like so far; +1; regards; and technically I would say the exact solution is closer to 0.512603 $\endgroup$ – J. W. Tanner Jul 9 '19 at 10:41
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    $\begingroup$ @TonyK. I never said that. I just gave a better approximation than $\frac 12$ which is clear from the plot. $\endgroup$ – Claude Leibovici Jul 9 '19 at 10:46
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    $\begingroup$ You said "the exact solution is $0.512603$", which is a rational number. $\endgroup$ – TonyK Jul 9 '19 at 10:48
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$X=\sqrt{1-x^2}$, replacing you have the equation $8X^8+32X^6-8X^4-80X^2-X+49=0$. It has $X=1$ and another positive ($<1$) solution.

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Expanding the left-hand side we get $$8\,{x}^{8}-64\,{x}^{6}+136\,{x}^{4}-32\,{x}^{2}+1$$ squaring this term and subtracting $(1-x^2)$

$${x}^{2} \left( 64\,{x}^{14}-1024\,{x}^{12}+6272\,{x}^{10}-17920\,{x}^{ 8}+22608\,{x}^{6}-8832\,{x}^{4}+1296\,{x}^{2}-63 \right) =0$$ by a numerical method we get the following solutions $$-.5126032239, -.3954020655, -.3272996816, 0., 0., .3272996816, .3954020655, .5126032239$$

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  • $\begingroup$ As a side note: Render $y=2x^2$ in the 14-degree factor. The resulting septic equation in $y$, multiplied by $2$ to clear a fraction, is $2$-Eisenstein irreducible. $\endgroup$ – Oscar Lanzi Jul 9 '19 at 10:01
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    $\begingroup$ THIS DOESN'T SOLVE THE GIVEN EQUATION Did you verify that the obtained numbers satisfy the given equation? I guess they (except $0$) make the left hand side negative, which is excluded. $\endgroup$ – user376343 Jul 9 '19 at 10:10
  • $\begingroup$ I have squared two times so additional solutions are possible, $\endgroup$ – Dr. Sonnhard Graubner Jul 9 '19 at 10:13

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