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?
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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. Any help please? For A just multiply both sides by 0.1-x and then subtract the sides since this is an equality, rather than inequality this is now of the form then 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.
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?
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 as that will be negative x will be positive if which it is, hence
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.
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