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Math Help - rearranging eqn to form quadratic eqn

  1. #1
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    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.

    Any help please?
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  2. #2
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    Quote Originally Posted by thatsnotalion View Post
    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

    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}

    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.
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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by thatsnotalion View Post
    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
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  5. #5
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    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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