Question

Oí

1) Resolva em lR :
$x^{4} - 5x^{2} - 4 = 0$​

Answer 1

Oie, Td Bom?!

$x {}^{4} - 5x {}^{2} - 4 = 0$

$x {}^{2 \: . \: 2} - 5x {}^{2} - 4 = 0$

$(x {}^{2} ) {}^{2} - 5x {}^{2} - 4 = 0$

• Substitua x² = t.

$t {}^{2} - 5t - 4 = 0$

• Coeficientes:

$a = 1 \: , \: b = - 5 \: ,\: c = - 4$

• Fórmula resolutiva:

$x = \frac{ - b± \sqrt{b{}^{2} - 4ac} }{2a}$

$t = \frac{ - ( - 5)± \sqrt{( - 5) {}^{2} - 4 \: . \: 1 \: . \: ( - 4) } }{2 \: . \: 1}$

$t = \frac{5± \sqrt{25 + 16} }{2}$

$t = \frac{5± \sqrt{41} }{2}$

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$t = \frac{5 + \sqrt{41} }{2}$

$t = \frac{5 - \sqrt{41} }{2}$

• Substitua t = x².

$x {}^{2} = \frac{5 + \sqrt{41} }{2} ⇒x {}^{2} = ± \frac{ \sqrt{10 + 2 \sqrt{41} } }{2}$

$x {}^{2} = \frac{5 - \sqrt{41} }{2} ⇒x∉\mathbb{R}$

• Solução:

$S = \left \{ - \frac{ \sqrt{10 + 2 \sqrt{41} } }{2} \: ,\: \frac{ \sqrt{10 + 2 \sqrt{41} } }{2} \right \}$

Att. Makaveli1996