Question

x em função de L em 8q/x^2=2q/(x-L)^2

Answer 1

Resposta:

Explicação passo-a-passo:

$\frac{8q}{x^{2}}=\frac{2q}{(x-L)^{2}}\\\\\frac{4}{x^{2}}=\frac{1}{x^{2}-2Lx+L^{2}}\\\\4x^{2}-8Lx+4L^{2}=x^{2}\\\\4x^{2}-x^{2}-8Lx+4L^{2}=0\\\\3x^{2}-8Lx+4L^{2}=0\\\\\\x=\frac{-(-8L)\pm\sqrt{(-8L)^{2}-4.3.4L^{2}}}{2.3}\\\\\\x=\frac{8L\pm\sqrt{64L^{2}-48L^{2}}}{2.3}\\\\\\x=\frac{8L\pm\sqrt{16L^{2}}}{2.3}\\\\\\x=\frac{8L\pm4L}{2.3}\\\\\\x=\frac{2(4L\pm2L)}{2.3}\\\\\\x=\frac{4L\pm2L}{3}\\\\\\x=\frac{4L+2L}{3}=\frac{6L}{3}=2L\\\\\\ou\\\\\\x=\frac{4L-2L}{3}=\frac{2L}{3}$

Answer 2

Olá rjm237...


X em função de L.

$\frac{8q}{ {x}^{2} } = \frac{2q}{ {(x - l)}^{2} } \\ \frac{4}{ {x}^{2} } = \frac{1}{ {(x - l)}^{2} } \\ 4 \times ( {x}^{2} - 2lx + {l}^{2} ) = 1 \times {x}^{2} \\ 3 {x}^{2} - 8lx + 4 {l}^{2} = 0$

Aplicando a fórmula de Bhaskara...

$x = \frac{ - ( - 8l) + - \sqrt{ {( - 8)}^{2} - 4 \times 3 \times 4 {l}^{2} } }{2 \times 3} \\ x = \frac{8l + - \sqrt{64 {l}^{2} - 48 {l}^{2} } }{6} \\ x = \frac{8l + - \sqrt{16 {l}^{2} } }{6} \\ x = \frac{8l + - 4l}{6} \\ \\ x1 = \frac{8l + 4l}{6} = 2l \\ x2 = \frac{8l - 4l}{6} = \frac{2l}{3}$

Resposta: x = 2L; x = 2L/3.


Espero ter ajudado!