Question

Qual é o valor de X para $48x^{4}$ + $16x^{2}$ = 64 ?

Answer 1

Resposta:

${48x}^{4} + {16x}^{2} = 64$

${48x}^{4} + 16 {x}^{2} - 64 = 0$

$( {48x}^{2} )^{2} + {16x}^{2} - 64 = 0$

${x}^{2} = y$

${48y}^{2} + 16y - 64 = 0$

$a = 48 \: \: \: b = 16 \: \: \: e \: \: \: c = - 64$

$x = \frac{ - ( + 16) \frac{ + }{ - } \sqrt{ {16}^{2} - 4 \times 48 \times ( - 64)} }{2 \times 48}$

$x = \frac{ - 16 \frac{ + }{ - } \sqrt{12544} }{96}$

$x = \frac{ - 16 \frac{ + }{ - } 112}{96}$

$x1 = \frac{ - 16 + 112}{96} = \frac{96}{96} = 1$

$x2 = \frac{ - 16 - 112}{96} = \frac{ - 128}{96} = \frac{4}{3}$

${x}^{2} = 1$

$x = \frac{ + }{ - } \sqrt{1}$

$x = \frac{ + }{ -} 1$

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

$x = \frac{ + }{ - } \sqrt{ \frac{4}{3} }$

$x = \frac{ + }{ - } \frac{2}{ \sqrt{3} }$

$x = \frac{ + }{ - } \frac{2 \sqrt{3} }{3}$

S = { 2\/3/3, - 2\/3/3, 1 , - 1}

Espero ter ajudado.