Rozwiąż równanie 2x⁴ - 3x² + 1 = 0.... Question from @Cetm

December 2018 2 19 Report

Rozwiąż równanie
2x⁴ - 3x² + 1 = 0

wik8947201 $\\2x^4-3x^2+1=0 \\ \\2x^4-2x^2-x^2+1=0 \\ \\2x^2(x^2-1)-(x^2-1)=0 \\ \\(x^2-1)(2x^2-1)=0 \\ \\(x+1)(x-1)(\sqrt2x+1)(\sqrt2-1)=0 \\ \\x=-1 \ \vee \ x=1 \ \vee \ x=-\frac{\sqrt2}{2} \ \vee \ x=\frac{\sqrt2}{2} \\ \\x\in\{-1, \ -\frac{\sqrt2}{2}, \ \frac{\sqrt2}{2}, \ 1\}$
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krysta $2x^4- 3x^2 + 1 = 0 \\ \\2x^4- 2x^2 -x^2+ 1 = 0 \\\\2x^2( x^2- 1) -(x^2- 1) = 0 \\\\ (x^2-1)(2x^2-1)=0\\ \\(x-1)(x+1)(x\sqrt{2}-1)(x\sqrt{2}+1)=0$

$x-1=0\ \ \ \vee \ \ \ x+1=0\ \ \ \vee \ \ \ x\sqrt{2}-1=0 \ \ \ \vee \ \ \ x\sqrt{2}+1 =0\\\\ x =1\ \ \ \vee \ \ \ x=-1 \ \ \ \vee \ \ \ x\sqrt{2}=1 \ \ \ \vee \ \ \ x\sqrt{2}=-1 \\\\x =1\ \ \ \vee \ \ \ x=-1 \ \ \ \vee \ \ \ x=\frac{1 }{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}} \ \ \ \vee \ \ \ x=-\frac{1}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}} \\\\ x =1\ \ \ \vee \ \ \ x=-1 \ \ \ \vee \ \ \ x=\frac{ \sqrt{2} }{2} \ \ \ \vee \ \ \ x=-\frac{\sqrt{2} }{2 }$

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