1. ## Using the Quadratic Formula Part 2

I was having difficulty solving a quadratic equation in which one has two different values for x as the numerator. Here is the problem:

(0.100 + X)(X)/(0.100 - X) = 1.8 x 10^-5

Any input on the above question will be greatly appreciated. Thank you to the individual who answered my last question.

2. Originally Posted by confused20
I was having difficulty solving a quadratic equation in which one has two different values for x as the numerator. Here is the problem:

(0.100 + X)(X)/(0.100 - X) = 1.8 x 10^-5

Any input on the above question will be greatly appreciated. Thank you to the individual who answered my last question.
Just like the last one you have to do a little work first.
$\frac{(0.1 + x)x}{0.1 - x} = 1.8 \times 10^{-5}$

Mulitply both sides by 0.1 - x:
$(0.1 + x)x = (1.8 \times 10^{-5})(0.1 - x)$

Now expand both sides:
$0.1x + x^2 = 1.8 \times 10^{-6} - (1.8 \times 10^{-5})x$

Get everything onto one sides of the equation:
$x^2 + (0.1 + 1 \times 10^{-5})x - 1.8 \times 10^{-6} = 0$

$x^2 + 0.10001x - 0.000018 = 0$