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Math Help - Is there a simpler solution?

  1. #1
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    Is there a simpler solution?

    Hi

    Relation
    \displaystyle{\frac{(500+x)}{700} = \frac{c}{40}}

    and
    \sqrt{c^2+40^2}=40+x

    c=\sqrt{x+80x}

    gives
    \displaystyle{\frac{(500+x)}{700}=\frac{\sqrt{x+8  0x}}{40}}

    How do I solve above?

    Is there a simpler solution?-image.gif
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  2. #2
    Member Pranas's Avatar
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    Quote Originally Posted by pelle2004 View Post
    Hi

    Relation
    \displaystyle{\frac{(500+x)}{700} = \frac{c}{40}}

    and
    \sqrt{c^2+40^2}=40+x

    c=\sqrt{x+80x}

    gives
    \displaystyle{\frac{(500+x)}{700}=\frac{\sqrt{x+8  0x}}{40}}

    How do I solve above?

    Click image for larger version. 

Name:	Image.gif 
Views:	11 
Size:	2.8 KB 
ID:	21115
    When dealing with real numbers, this logic might be handy:

    \displaystyle \[a = \sqrt b  \Rightarrow \left\{ \begin{gathered}<br />
  {a^2} = b \hfill \\<br />
  b \geqslant 0 \hfill \\<br />
  a \geqslant 0 \hfill \\ <br />
\end{gathered}  \right.\]
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by pelle2004 View Post
    Hi

    Relation
    \displaystyle{\frac{(500+x)}{700} = \frac{c}{40}}

    and
    \sqrt{c^2+40^2}=40+x

    c=\sqrt{x+80x}

    gives
    \displaystyle{\frac{(500+x)}{700}=\frac{\sqrt{x+8  0x}}{40}}

    How do I solve above?

    Click image for larger version. 

Name:	Image.gif 
Views:	11 
Size:	2.8 KB 
ID:	21115
    Square both sides to get a quadratic. Solve the quadratic and check the solutions by substituting back into the above to make sure you have not introduced spurious solutions by squaring.

    CB
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  4. #4
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    Not shure what I should do by 'quadratic'

    \displaystyle{\frac{500+x}{17,5}}=\sqrt{x+80x}

    \displaystyle{\frac{500+x+1000x}{17,5}}=x+80x

    x+80x-\displaystyle{\frac{500}{17,5}-\frac{x}{17,5}-\frac{1000}{17,5}x}=0

    (17,5-1)x+(17,5*80-1000)x-500=0

    305,25x+23500x-250000=0

    x+\displaystyle{\frac{23500}{305,25}}x=250000

    I guess....

    y(x)=\frac{250000*305,25}{23500}+Ce^{-\frac{23500}{305,25}x}

    But then I need some help to be able to solve x for y=0 ...
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  5. #5
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    earboth's Avatar
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    Quote Originally Posted by pelle2004 View Post
    Not shure what I should do by 'quadratic'

    \displaystyle{\frac{500+x}{17,5}}=\sqrt{x+80x}

    \displaystyle{\frac{500+x+1000x}{17,5}}=x+80x

    x+80x-\displaystyle{\frac{500}{17,5}-\frac{x}{17,5}-\frac{1000}{17,5}x}=0

    (17,5-1)x+(17,5*80-1000)x-500=0

    305,25x+23500x-250000=0
    The last quoted line is OK.

    A quadratic equation

    a x^2+bx +c = 0 has the solution x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

    you know: a = 305.25, b = 23500, c = -250000

    so you can calculate the value of x. (Btw the negative solution isn't very plausible with your question)
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  6. #6
    MHF Contributor
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    Quote Originally Posted by pelle2004 View Post
    \displaystyle{\frac{(500+x)}{700}=\frac{\sqrt{x+8  0x}}{40}}
    Less confusion if next is rearranging this way:

    35SQRT(x^2 + 80x) = 2x + 1000 ; then:

    1225x^2 + 98000x = 4x^2 + 4000x + 1000000

    1221x^2 + 94000x - 1000000 = 0
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  7. #7
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    Quote Originally Posted by earboth View Post
    The last quoted line is OK.

    A quadratic equation

    a x^2+bx +c = 0 has the solution x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

    you know: a = 305.25, b = 23500, c = -250000

    so you can calculate the value of x. (Btw the negative solution isn't very plausible with your question)
    x=9.4727287939

    /Thanks!!
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