Question

${3x}^{2} = \frac{x}{4} + \frac{1}{64}$
a diferença entre a maior raiz e a menor raiz vale:
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Answer 1

$3 {x}^{2} = \frac{x}{4} + \frac{1}{64} \: \: \: \: \: \times (64) \\ 192 {x}^{2} = 16x + 1 \\ 192 {x}^{2} - 16x - 1 = 0 \\ a = 192 \\ b = - 16 \\ c = - 1 \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4 \times a \times c } }{2 \times a} \\ x = \frac{ - ( - 16) + - \sqrt{ {( - 16)}^{2} - 4 \times 192 \times ( - 1) } }{2 \times 192} \\ x = \frac{16 + - \sqrt{256 + 768} }{384} \\ x = \frac{16 + - \sqrt{1024} }{384} \\ x = \frac{16 + - 32}{384} \\ x1= \frac{16 + 32}{384} = \frac{48}{384} = \frac{1}{8} </p><p></p><p>x2 = \frac{16 - 32}{384} = \frac{ - 16}{384} = \frac{ - 1}{24} \\ diferenca = \\ \frac{1}{8} - ( - \frac{1}{24} ) \\ \frac{1}{8} + \frac{1}{24} \\ \frac{1}{8} \times (3) = \frac{3}{24} \\ \frac{3}{24} + \frac{1}{24} = \frac{4}{24} = \frac{1}{6}$