Question

Equações Biquadrada.

A)
(2 {x}^{2} - 5) {}^{2} = 10(2 {x}^{2} - 5) + 39[/tex]
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Answer 1

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$\sf(2x^2-5)^2=10\cdot\!(2x^2-5)+39\\\sf 4(x^2)^2-20x^2+25=20x^2-50+39\\\sf 4(x^2)^2-20x^2-20x^2+25+50-39=0\\\sf 4(x^2)^2-40x^2+36=0\div4\\\sf (x^2)^2-10x^2+9=0\\\rm fac_{\!\!,}a~x^2=t\\\sf t^2-10t+9=0\\\sf s=-\dfrac{b}{a}=-\dfrac{-10}{1}=10\\\sf p=\dfrac{c}{a}=\dfrac{9}{1}=9\\\sf dois~n\acute umeros~cuja~soma~\acute e~10~e~o~produto~\acute e~9\\\sf s\tilde ao~ 9~e~1~pois~9+1=10~e~9\cdot1=9 \\\sf portanto~estas~s\tilde ao~as~ra\acute izes.da\acute i:\\\sf x^2=t\\\sf x^2=9\\\sf x=\pm\sqrt{9}\\\sf x=\pm3$

$\sf x^2=t\\\sf x^2=1\\\sf x=\pm\sqrt{1}\\\sf x=\pm1\\\huge\boxed{\boxed{\boxed{\boxed{\sf S=\{-3,-1,1,3\}\checkmark}}}}$