Question

Resolva a equação 2x⁴ - 20x² - 12 = 0​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{2x^4 - 20x^2 - 12 = 0}$

$\mathsf{y = x^2}$

$\mathsf{2y^2 - 20y - 12 = 0}$

$\mathsf{y^2 - 10y - 6 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (-10)^2 - 4.1.(-6)}$

$\mathsf{\Delta = 100 + 24}$

$\mathsf{\Delta = 124}$

$\mathsf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{10 \pm \sqrt{124}}{2} \rightarrow \begin{cases}\mathsf{y' = \dfrac{10 + 2\sqrt{31}}{2} = 5 + \sqrt{31}}\\\\\mathsf{y'' = \dfrac{10 - 2\sqrt{31}}{2} = 5 - \sqrt{31}}\end{cases}}$

$\mathsf{x^2 = y'}$

$\mathsf{x' = \pm\: \sqrt{5 + \sqrt{31}}}$

$\mathsf{x^2 = y''}$

$\mathsf{x'' = \pm\: \sqrt{5 - \sqrt{31}}}$

$\boxed{\boxed{\mathsf{S = \{\sqrt{5 + \sqrt{31}};-\sqrt{5 + \sqrt{31}};\sqrt{5 - \sqrt{31}};-\sqrt{5 - \sqrt{31}}\}}}}$

Answer 2

$2x {}^{4} - 20x {}^{2} - 12 = 0 \\ \boxed{t ⟷x {}^{2} } \\ 2t {}^{2} - 20t - 12 = 0 \\ \frac{2t {}^{2} - 20t - 12 }{2} = 0 \\ t {}^{2} - 10t - 6 = 0$

$\boxed{a = 1 \: ,\: b = - 10 \: , \: c = - 6}$

$x = \frac{ - b± \sqrt{b {}^{2} - 4ac } }{2a} \\t= \frac{ - ( - 10)± \sqrt{( - 10){}^{2} - 4 \: . \: 1 \: . \: ( - 6)} }{2 \: . \: 1} \\ t = \frac{10± \sqrt{100 + 24} }{2} \\ t = \frac{10± \sqrt{124} }{2} \\ t = \frac{10±2 \sqrt{31} }{2} \\ t = \frac{10 +2 \sqrt{31} }{2} = \boxed{5 + \sqrt{31} } \\ t = \frac{10 - 2 \sqrt{31} }{2} = \boxed{5 - \sqrt{31} }$

$\boxed{t ⟷x {}^{2} } \\ x {}^{2} = 5 + \sqrt{31} ⟶ \boxed{x = ± \sqrt{5 + \sqrt{31} }} \\ x {}^{2} = 5 - \sqrt{31} ⟶ \boxed{ x∉\mathbb{R}}$

$\boxed{S = \left \{ - \sqrt{5 + \sqrt{31} } \: , \: \sqrt{5 + \sqrt{31} } \right \}}$

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