# Biar Rame

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Solve the equation

$x^3-3x=\sqrt{x+2}$

[Problem 24 - 101 Problems in Algebra]

Edited by BeingNotknown Ya

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Batas nilai $x \ge -2$.

\begin{align*} x^3-3x &=\sqrt{x+2} \\ (x^3 - 3x)^2 &= x+2 \\ x^6 + 9x^2 - 6x^4 -x-2 & = 0 \\ x^6 -6x^4 + 9x^2 - x - 2 &=0 \end{align*}

Terhenti sampai sini  + gak ada kertas oret-oretan. Mnurutku pakai dalil vietta selanjutnya.

Edited by Wildan Bagus W

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