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Simplifying an algebra question

  1. Dec 31, 2012 #1
    Hello,

    Ive was reading my literature today and it showed me some algebra, which i did not understand how it went from one step to the next...

    What was done to simplify
    V= x^2 ( 75-x^2 / 2x ) into V= x/2 ( 75-x^2 )

    & then what was done to simplify ^^^

    V= x/2 ( 75-x^2 ) into V= 1/2 ( 75x-x^3 )

    (for anyone interested the above has now been differentiated)

    I would now like to know when finding x how they go from
    1/2 ( 75-3x^2 ) = 0 to this 75 = 3x^2



    Thanks for any help. I know I'm missing something simple here?
     
  2. jcsd
  3. Dec 31, 2012 #2

    HallsofIvy

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    Staff Emeritus
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    You mean V= x^2((75- x^2)/(2x)). That is, the entire "75- x^2" is divided by 2x. You can factor 1/(2x) out to get (x^2/2x)(75- x^2)= (1/2)(x)(75- x^2).

    Now multiply the "x" back into the 75- x^2: x/2= (1/2)(x)(75- x^2)= (1/2)(75(x)- x^2(x))= (1/2)(75x- x^3).

    Multiply both sides by 2 to get 75- 3x^2= 0
    then add 3x^2 to both sides: 75= 3x^2



    Although you dont mention it, an obvious next thing to do is to divide both sides by 3:
    25= x^2. What you do now depends upon what the question is! If it is to solve for x, take the square root of both sides.
     
  4. Jan 1, 2013 #3
    Ok, thanks for taking the time to reply, just to let you know it can sometimes take time for things to "click in my head" lol.. Im still confused, so can I break it down and show you where Im at...




    I understand the last bit as you have cancelled out one of the x's from the top and bottom of the fraction.



    Im confused as to where the part in red 1/(2x) came from?
    I also dont understand why the 2x disappears from under the 75-x^2 and then appears under the x^2?


    Thanks
     
    Last edited: Jan 1, 2013
  5. Jan 1, 2013 #4
    It's like this:

    ##V= x^2 (\frac{75-x^2}{2x})=\frac{1}{2x}x^2(75-x^2)=\frac{x}{2}(75-x^2)##
     
  6. Jan 1, 2013 #5

    I like Serena

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    Homework Helper

    Here's a couple of the rules for working with fractions and multiplications:
    $$\frac 2 3 = 2 \cdot \frac 1 3$$
    $$2 \cdot 3 \cdot 4 = 4 \cdot 2 \cdot 3$$
    $$\frac {3 \cdot 2} {5 \cdot 2} = \frac 3 5$$
    These are the rules MrWarlock applied.
     
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