What is x if $x - \frac{3}{x^{2}} + 14 = 8$ $x-3/x^2+14=8$ ?

Question

1726 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-14T04:52:04

$x - \frac{3}{x^{2}} + 14 = 8$ $x-3/x^2 + 14 =8$

Subtract 8 from each side:
$x - \frac{3}{x^{2}} + 6 = 0$ $x - 3/x^2 + 6 = 0$

Multiply each side by $x^{2}$ $x^2$ :
$(x^{2}) (x - \frac{3}{x^{2}} + 6) = 0$ $(x^2)(x - 3/x^2 + 6) = 0$

Distribute and simplify:
$x^{3} - 3 + 6 x^{2} = 0$ $x^3 - 3 + 6x^2 =0$

$0 = x^{3} + 6 x^{2} - 3$ $0 = x^3 + 6x^2 - 3$

From here, I think the best option to solve this would be to use a graphing calculator. On the TI-84 plus, I used the numeric solver.

$x = - .671, x = .756, x = 5.914$ $x = -.671, x=.756, x=5.914$

What is x if x−3x2+14=8x-3/x^2+14=8?