Chemical Quadratics

Algebra Level 3

$\frac {x^{2}} {(0,01-x)} = 6,3 \times 10^{-4}$

Solve for positive x with 4 decimal places

1 solution

Ebo Peerbooms
Mar 3, 2016

First write out the equation in quadratic form:

$x^{2} + 6,3 \times 10^{-4} x \times 6,3 \times 10^{-6} = 0$

Apply the quadratic formula:

$\frac{-6,3 \times 10^{-4} \frac{+}{-} \sqrt{6,3 \times 10^{-4} + 4 \times 1 \times 6,3 \times 10^{-6}}}{2} = 0,0022$