I"m supposed to solve this equation. It"s indigenous a jamesmerse.com challenge so addressing it through hand would be preferable (no quartic formulas).I thought about making $u = x^2-3x-2$ obviously but it leads to one more quartic equation. I also tried the substitution $u=x+2$, and after the totality expand trinomial, simplify, invoke rational source theorem and also test roots, i still acquired nothing out of it.

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I noticed that $x^2-3x-2$ can"t be factored unique so ns dunno what other route come take. Numerous equations ns tackled in jamesmerse.com contests might make use of nice trigonometric substitutions, yet none in particular pop in my head right now.

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


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

Note that any type of solution come $x = x^2 - 3x - 2$ is additionally a equipment to the offered equation.

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

Alternatively, see this virtually 10 year old subject on the arts of difficulty Solving Forum, or this slightly newer thread i beg your pardon discusses a similar problem.


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