I"m supposed to settle this equation. It"s native a bsci-ch.org contest so resolving it through hand would be preferable (no quartic formulas).I thought around making $u = x^2-3x-2$ obviously however it leader to one more quartic equation. I likewise tried the substitution $u=x+2$, and after the totality expand trinomial, simplify, invoke rational root theorem and test roots, i still obtained nothing out of it.
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I noticed the $x^2-3x-2$ can"t it is in factored nicely so ns dunno what various other route to take. Lots of equations i tackled in bsci-ch.org contests can make usage of pretty trigonometric substitutions, however none in specific pop in my head ideal now.
If everyone can give me clues or a full solution, that would be awesome. Thanks!
Since you have a quartic, over there are potentially 4 real solutions.
Note that any solution to $x = x^2 - 3x - 2$ is also a solution to the offered equation.
Hence, $x^2 - 4x - 2$ is a aspect of the quartic. Now uncover the other factor, and solve both quadratics.
Alternatively, watch this nearly 10 year old thread on the art of trouble Solving Forum, or this slightly more recent thread which discusses a comparable problem.
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settle $sqrt1 + sqrt1-x^2left(sqrt(1+x)^3 + sqrt(1-x)^3 ight) = 2 + sqrt1-x^2 $
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