<s>[INST] How to solve an irrational equation? I want to solve this equation
$$2 (x-2) \sqrt{5-x^2}+(x+1)\sqrt{5+x^2} = 7 x-5.$$
I tried
The given equation equavalent to
$$2 (x-2) (\sqrt{5-x^2}-2)+(x+1)(\sqrt{5+x^2}- 3)=0$$
or
$$(x-2)(x+1)\left [\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2}\right ] = 0.$$
I see that, the equation
$$\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2} = 0$$
has unique solution $x = 2$, but I can not solve. How can I solve this equation or solve the given equation with another way?
 [/INST] The usual way to do something like this is to square to get rid of one radical, rearrange, and square again to get rid of the remaining radical, as @vhspdfg says in a comment. I won’t detail the intermediate steps, but the octic polynomial I got was
$$
 2000 - 1600x - 3480x^2 + 2960x^3 + 825x^4 - 1340x^5 + 510x^6 - 140x^7 + 25x^8\\
=5(x+1)^2(x-2)^3(5x^3  - 8x^2 + 65x - 50)\,.
$$
The cubic seems to have no rational roots, and if this is correct, it’s irreducible.
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