1. ## rearranging eqn to form quadratic eqn

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.

2. Originally Posted by thatsnotalion
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.

For A

$k=\frac{x^2}{0.1-x}$

just multiply both sides by 0.1-x and then subtract the sides since this is an equality, rather than inequality

$x^2=(0.1-x)k=0.1k-xk$

$x^2+kx-0.1k=0$

this is now of the form

$ax^2+bx+c=0$

then

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

b=k
c=-0.1k

substitute these to find x in terms of k
and use that k to answer part B.

3. thanks for that. i figured out the answer to part a as i waited however i keep getting weird answers when i put 1.75*10^-5 in the eqn?

4. Originally Posted by thatsnotalion
thanks for that. i figured out the answer to part a as i waited however i keep getting weird answers when i put 1.75*10^-5 in the eqn?
If x must be positive, then we can forget about

$x=\frac{-k-\sqrt{k^2+0.4k}}{2}$

as that will be negative

x will be positive if

$\sqrt{k^2+0.4k}>k$

which it is, hence

$x=\frac{-0.0000175+\sqrt{(0.0000175)^2+0.4(0.0000175)}}{2}$

$=\frac{-0.0000175+\sqrt{0.00000700031}}{2}=\frac{-0.0000175+0.002646}{2}=\frac{0.0026285}{2}=0.00131$

5. thankyou so much. i was doing it bit by bit and kept getting a negative number under the square root sign which put me off.

thanks again.