I"m supposed to settle this equation. It"s indigenous a bromheads.tv contest so resolving it by hand would certainly be preferable (no quartic formulas).I thought about making $u = x^2-3x-2$ obviously yet it leads to another quartic equation. I likewise tried the substitution $u=x+2$, and after the whole expand trinomial, simplify, invoke rational source theorem and test roots, i still obtained nothing the end of it.

You are watching: (3x^2)^2

I noticed the $x^2-3x-2$ can"t be factored unique so i dunno what other route to take. Numerous equations ns tackled in bromheads.tv contests can make usage of pretty trigonometric substitutions, yet none in details pop in mine head best now.

If anyone can give me clues or a complete solution, that would certainly be awesome. Thanks!


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Since you have a quartic, over there are potentially 4 actual solutions.

Note that any solution to $x = x^2 - 3x - 2$ is also a systems to the given equation.

Hence, $x^2 - 4x - 2$ is a variable of the quartic. Now uncover the other factor, and solve both quadratics.

Alternatively, see this practically 10 year old subject on the art of difficulty Solving Forum, or this slightly newer thread i m sorry discusses a similar problem.


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