# Solve the following equation : $2(2(x^{2}-2)^{2}-7)^{2}-1=\sqrt{1-x^2}$

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?

• I agree squaring can introduce extraneous solutions, but I disagree that $0$ is the only solution; see the answer by Claude Leibovici – J. W. Tanner Jul 9 '19 at 10:32

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 $$\approx0.512603$$.

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 $$x=\frac 12+\frac{6954 \sqrt{3}-7209}{310977+42536 \sqrt{3}}\approx 0.512572$$

• By the symmetry you mentioned, if $x$ is a root, then so is $-x$ – J. W. Tanner Jul 9 '19 at 10:28
• @J.W.Tanner. For sure ! I supposed (may be wrongly) that it was explicit. – Claude Leibovici Jul 9 '19 at 10:38
• 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 – J. W. Tanner Jul 9 '19 at 10:41
• @TonyK. I never said that. I just gave a better approximation than $\frac 12$ which is clear from the plot. – Claude Leibovici Jul 9 '19 at 10:46
• You said "the exact solution is $0.512603$", which is a rational number. – TonyK Jul 9 '19 at 10:48

$$X=\sqrt{1-x^2}$$, 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.

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 $$-.5126032239, -.3954020655, -.3272996816, 0., 0., .3272996816, .3954020655, .5126032239$$

• 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. – Oscar Lanzi Jul 9 '19 at 10:01
• 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. – user376343 Jul 9 '19 at 10:10
• I have squared two times so additional solutions are possible, – Dr. Sonnhard Graubner Jul 9 '19 at 10:13