Question

Como resolver essa equação biquadratica passo a passo
$(x {}^{2} - 2) {}^{2} + (x {}^{2} - 4) {}^{2} = 4$

$x {}^{4} - \frac{x {}^{2} - 3 }{2} = \frac{2x {}^{2} + 4 }{3}$
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Answer 1

$(x {}^{2} - 2) {}^{2} + (x {}^{2} - 4) {}^{2} = 4 \\ {x}^{4} - 4 {x}^{2} + 4 + {x}^{4} - 8 {x}^{2} + 16 - 4 \\ = 0$

$2{x}^{4} - 12 {x}^{2} + 16 = 0 \div 2 \\ {x}^{4} - 6 {x}^{2} + 8 = 0$

Fazendo

${x}^{2}=t$

${ ({x}^{2})}^{2} - 6 {x}^{2} + 8 = 0 \\ {t}^{2} - 6t + 8 = 0$

$a=1\\b=-6\\c=8$

$\Delta={b}^{2}-4ac\\\Delta={(-6)}^{2}-4.1.8\\\Delta=36-32=4$

$t=\dfrac{-b\pm\sqrt{\Delta}}{2a}$

$t=\dfrac{-(-6)\pm\sqrt{4}}{2.1}\\t=\dfrac{6\pm2}{2}$

$t_{1}=\dfrac{6+2}{2}=\dfrac{8}{2}=4\\t_{2}=\dfrac{6-2}{2}=\dfrac{4}{2}=2$

${x}^{2}=t\\{x}^{2}=4\\x=\pm\sqrt{4}\\x=\pm2$

${x}^{2}=t\\{x}^{2}=2\\x=\pm\sqrt{2}$

$s=\{ - 2, - \sqrt{2} ,\sqrt{2},2\}$

$x {}^{4} - \frac{x {}^{2} - 3 }{2} = \frac{2x {}^{2} + 4 }{3}$

Multiplicando por 6 dos dois lados temos

$6 {x}^{4} - 3( {x}^{2}-3) = 2( 2{x}^{2} + 4) \\ 6 {x}^{4}-3 {x}^{2} + 9 = 4{x}^{2} + 8$

$6 {x}^{4}-3{x}^{2} - 4 {x}^{2} + 9 - 8 = 0$

$6 {x}^{4} - 7 {x}^{2} + 1 = 0 \\ 6{ ({x}^{2} )}^{2} - 7 {x}^{2} + 1 = 0$

fazendo

${x}^{2}=t$

$6 {t}^{2} - 7t + 1 = 0$

$a=6\\b=-7\\c=1$

$\Delta={b}^{2}-4ac\\\Delta={(-7)}^{2}-4.6.1\\\Delta=49-24=25$

$\huge \: t=\dfrac{-b\pm\sqrt{\Delta}}{2a}$

$\huge \: t=\dfrac{-( - 7)\pm\sqrt{25}}{2.6}$

$\huge \: t=\dfrac{7\pm5}{12}$

$\huge \: t_{1}=\dfrac{7+5}{12}=1$

$\huge t_{2}=\dfrac{7-5}{12}=\dfrac{1}{6}$

$\huge{x}^{2}=1\\ \huge \\ \huge x=\pm\sqrt{1}\\ \huge \: x=\pm1$

$\huge \: {x}^{2}=\dfrac{1}{6}$

$\huge x=\pm\sqrt{\dfrac{1}{6}} = \dfrac{ \sqrt{6} }{6}$