I"m claimed to solve this equation. It"s from a dearteassociazione.org dispute so resolving it by hand would certainly be preferable (no quartic formulas).I thought about making \$u = x^2-3x-2\$ obviously yet it leader to one more quartic equation. I additionally tried the substitution \$u=x+2\$, and also after the totality expand trinomial, simplify, invoke rational source theorem and also test roots, i still got nothing the end of it.

I noticed that \$x^2-3x-2\$ can"t it is in factored unique so i dunno what various other route to take. Several equations ns tackled in dearteassociazione.orgematics contests might make usage of nice trigonometric substitutions, yet none in details pop in mine head right now.

If everyone can provide me hints or a complete solution, that would certainly be awesome. Thanks!

Since you have actually a quartic, there are perhaps 4 genuine solutions.

Note that any kind of solution to \$x = x^2 - 3x - 2\$ is additionally a equipment to the given equation.

Hence, \$x^2 - 4x - 2\$ is a aspect of the quartic. Now find the various other factor, and also solve both quadratics.

Alternatively, check out this almost 10 year old subject on the arts of problem Solving Forum, or this slightly more recent thread i beg your pardon discusses a similar problem.

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solve \$sqrt1 + sqrt1-x^2left(sqrt(1+x)^3 + sqrt(1-x)^3 ight) = 2 + sqrt1-x^2 \$

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