Solving an equation in $\mathbb{R}$

Question

Let $a>0$. I want to solve the following equation: $$x^2 -2x +\sqrt{1-x^2} -\left(\frac{1}{a^2}+1\right) = 0.$$

I did a plot of the function on Matlab and I see that for $a=1,2,2000,...$ etc, and I see that the function has always a solution over $[-1,1]$.

I tried the following variable change: $u= \sqrt{1-x^2}$ however, it did not work (because we obtain the same equation).

Do you want to find some explicit solution or show the existence of a solution? — Infinity_hunter, Oct 03 '22 at 12:21
The equation does have explicit solutions, but they are fairly complicated. — L. F., Oct 03 '22 at 12:25
@L.F. that's what I found using MAPLE. A very complicated (not explicit one) — hanava331, Oct 03 '22 at 12:34
What I sent you is an explicit solution in mathematical terms. Complicatedness has nothing to do with explicitness. Moreover, I consider it highly likely that there are no (fundamentally) simpler forms. — L. F., Oct 03 '22 at 12:39
WolframAlpha gives explicit solution, it looks complicated because it is expression of kind $x_i=f(a)$. Of course this solution can be written in shorter form with separating of repeating parts like $x_i=g(d,e)$, where $d=h(a)$, $e=i(a)$. — Ivan Kaznacheyeu, Oct 03 '22 at 12:51

score 1 · Answer 1 · answered Oct 03 '22 at 12:35

Let $c= 1+ \frac{1}{a^2}$ then your equation can be written as $$x^2-2x-c= -\sqrt{1-x^2}$$ By squaring both sides we get $$ x^4 - 4 x^3 - 2 c x^2 + 5 x^2 + 4 c x + c^2 - 1=0$$ which is a quartic equation and here you can see how to solve them.

To determine if the roots are real use the properties of discriminant $Δ = 256 c^4 - 128 c^3 - 1520 c^2 - 1536 c - 848$.

score 0 · Answer 2 · answered Oct 03 '22 at 12:44

0

Let $X=x-1$ and $b=2+\frac1{a^2}$, solve the quartic equation (in $[-2,0]$) $$X^4+(1-2b)X^2+2X+b^2=0.$$

answered Oct 03 '22 at 12:44

Anne Bauval

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Solving an equation in $\mathbb{R}$

2 Answers2