Hi there. i am having trouble solving this question. it has 2 parts.
PART A. rearrange k=(x^2)/(0.1-x) to obtain a quadratic eqn for x
PART B. when k=1.75 x 10^-5, find x. x cannot be negative.
Any help please?
For A
$\displaystyle k=\frac{x^2}{0.1-x}$
just multiply both sides by 0.1-x and then subtract the sides since this is an equality, rather than inequality
$\displaystyle x^2=(0.1-x)k=0.1k-xk$
$\displaystyle x^2+kx-0.1k=0$
this is now of the form
$\displaystyle ax^2+bx+c=0$
then
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Your a=1
b=k
c=-0.1k
substitute these to find x in terms of k
and use that k to answer part B.
If x must be positive, then we can forget about
$\displaystyle x=\frac{-k-\sqrt{k^2+0.4k}}{2}$
as that will be negative
x will be positive if
$\displaystyle \sqrt{k^2+0.4k}>k$
which it is, hence
$\displaystyle x=\frac{-0.0000175+\sqrt{(0.0000175)^2+0.4(0.0000175)}}{2}$
$\displaystyle =\frac{-0.0000175+\sqrt{0.00000700031}}{2}=\frac{-0.0000175+0.002646}{2}=\frac{0.0026285}{2}=0.00131$