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Math Help - Simple Equation Help

  1. #1
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    Question Simple Equation Help

    Hey All,

    I am looking through an engineering book and it has assumed the user can solve this equation and given the answer for C, but I cant figure out how to solve it myself. Can anyone help??

    4c2+1820c+17400 = 238900

    The 2 in 4c2 means squared.

    The books answer is C = 99.8

    Hope you can help!

    Thanks
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  2. #2
    Senior Member yeKciM's Avatar
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    c_1 \approx 99,8

    c_2 \approx -554.8

    u can use 2 ways ... quadratic formula .... or completing the square ...
    Last edited by yeKciM; July 24th 2010 at 04:37 AM. Reason: Sorry... did =23980... now its corrected .. :D
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  3. #3
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    Thanks yeKciM, I will look up quadratic formula... although im not sure how you have solved both of the c's? I am just trying to make C be on its own.

    Thanks!
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  4. #4
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    Use "^" to get powers in LaTex.
    Although it is not necessary, I would recommend dividing each number by 4 to simplify the equation: c^2+ 455c+ 4350= 59725. Now, since the "standard" form for a quadratic equation is " ax^2+ bx+ c= 0, subtract 59725 from each side to get c^2+ 455c- 55375= 0

    There are several ways to solve quadratic equations but most general is the "quadratic formula":
    solutions to ax^2+ bx+ c= 0 are given by \frac{-b\pm\sqrt{b^2- 4ac}}{2a}. Note the " \pm". There are two solutions to this equation- although there may be other reasons why a negative solution is not valid.
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  5. #5
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    I understand the first bit where we get rid of 59725 and make it look a little simpler, but then that equations just kind of blows me away. Is that the answer or is that what I need to do to get the answer. Do I change a and b to 455 and -553775? and C is still C?

    Thanks!
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  6. #6
    Senior Member yeKciM's Avatar
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    a=1 , b= 455 , c=-55375

    ur  C is  x in  ax^2+bx+c=0

    here u have commands for LaTex... so u can write it little better
    LaTeX:Commands - AoPSWiki
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  7. #7
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    Woohoo! 99.8.

    Thanks for helping me understand guys, that's awesome!
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  8. #8
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    Question

    Ok, I have just tried it on another one but I don't think my answer is right. Could someone check it please?


    4c^2+1490c+9310=88889
    c^2+1490c+9310=2222.25
    c^2+1490c+7087.75=0
    -1490+\sqrt{1490^2-4x1x7087.75}
    / 2X1
    = -4.77
    Last edited by Gazeranco; July 24th 2010 at 05:28 AM. Reason: untidy!
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  9. #9
    Senior Member yeKciM's Avatar
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    heheheheh i guess that x-es are *

    solution for that one .... is

    c_1\approx47.38 , c_2\approx-419.88

    4c^2+1490c+9310=88889

    4c^2+1490-79579=0

    so u have that a=4, b=1490, c=-79579

    u put it in...

    C_{1/2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
    Last edited by yeKciM; July 24th 2010 at 05:42 AM.
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  10. #10
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    Qudratic is:-

    c^2 + 1490c + 7087.75 = 0
    Last edited by Gazeranco; July 24th 2010 at 05:34 AM. Reason: wrong
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  11. #11
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    I see, so I need the A for it to work. But in the last example we used 1 as A. Is that right? Why does that not work in this instance. 47.38 sounds more like it!

    I see we now also have:-

    C X 0.5

    Is this how you removed the C from 1490?
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  12. #12
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Gazeranco View Post
    Ok, I have just tried it on another one but I don't think my answer is right. Could someone check it please?


    4c^2+1490c+9310=88889
    c^2+1490c+9310=2222.25
    c^2+1490c+7087.75=0
    -1490+\sqrt{1490^2-4x1x7087.75}
    / 2X1
    = -4.77
    dude... there's no need for u to divide with 4... but when u do that u divide all .... meaning that u can't from
    4c^2+1490c+9310=88889 divide with 4 get what u got
    4c^2+1490c+9310=2222.25 is wrooooong

    if u are gonna divide it u divide all so u get
    c^2+372.5c+2327.5=22222.25



    P.S. basic
    equation would not be changed if u divide it or multiply it with number different from zero... (but all elements of equation)
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  13. #13
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    lol I see... Doh!

    Still not sure where the half came from in your solution, could explain that part please?
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  14. #14
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Gazeranco View Post
    I see, so I need the A for it to work. But in the last example we used 1 as A. Is that right? Why does that not work in this instance. 47.38 sounds more like it!

    I see we now also have:-

    C X 0.5

    Is this how you removed the C from 1490?

    wait...
    let's go easy
    when u got any quadratic equation... any u have
    ax^2+bx+c=0 now in ur equations that C is actually x or unknown ? are we ok till here ?

    now when u look quadratic formula to solve quadratic equation like this

    x_{1/2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    u will have solution of quadratic equation .... but these a,b and c that u put in formula are : a is number that is with x^2, when it is let's say [x^2] means that a=1 ... means it's 1*x^2.... b is number that stands with just x , and c is number that have no x-es
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  15. #15
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Gazeranco View Post
    lol I see... Doh!

    Still not sure where the half came from in your solution, could explain that part please?

    lol that's not C*0,5 it's C one two ... meaning there are two solutions .... witch u get from \pm if there exist sqrt... if there is not sqrt meaning its 0 u have only one solution

    like this

    C_1=\frac{-b+\sqrt{b^2-4ac}}{2a}

    C_2=\frac{-b-\sqrt{b^2-4ac}}{2a}
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