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Math Help - Rearrange in terms of x?

  1. #1
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    Rearrange in terms of x?

    Please could someone help me with this?

    Is it possible to rearrange the following equation so that it is expressed in terms of x? Have been trying for ages without luck.

    x^2 = b^2 + (c - (ac/(a+x)))^2

    Any help would be much appreciated.
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  2. #2
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    Apologies if I have posted this in the wrong place.
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  3. #3
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    x^2 = b^2 + \left(c - \dfrac{ac}{a+x}\right)^2

    A good place to start would be to use foil to expand that squared term

    \left(c - \dfrac{ac}{a+x}\right)^2 = c^2 - \dfrac{2ac^2}{a+x} + \dfrac{a^2c^2}{(a+x)^2}

    You should then be able to multiply through by (a+x)^2 to clear the fraction

    x^2(a+x)^2 = b^2(a+x)^2 + c^2(a+x)^2 - 2ac^2(a+x) + a^2c^2


    You will eventually have to expand it and it won't be pretty but it should be simple enough using FOIL and the distributive property
    Last edited by e^(i*pi); January 22nd 2011 at 03:12 AM. Reason: suggested multiplying by (a+x)^2 instead of its expanded form
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  4. #4
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    Thanks for that. I'll have a go but think I've bitten off more than I can chew with this one. My mathematical capabilites are not great.
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  5. #5
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    Quote Originally Posted by EdCoventry View Post
    Please could someone help me with this?

    Is it possible to rearrange the following equation so that it is expressed in terms of x? Have been trying for ages without luck.

    x^2 = b^2 + (c - (ac/(a+x)))^2

    Any help would be much appreciated.

    \displaystyle{x^2=b^2+\left(c-\frac{ac}{a+x}\right)^2\Longrightarrow x^2-\left(c-\frac{ac}{a+x}\right)^2=b^2\Longrightarrow}

    \displaystyle{\Longrightarrow \left(x-\left(c-\frac{ac}{a+x}\right)\right)\left(x+\left(c-\frac{ac}{a+x}\right)\right)=b^2\Longrightarrow}

    \displaystyle{\Longrightarrow\left(x-\frac{cx}{a+x}\right)\left(x+\frac{cx}{a+x}\right)  =b^2\Longrightarrow \left(x^2+(a-c)x\right)\left(x^2+(a+c)x\right)=b^2x+ab^2}

    \displaystyle{\Longrightarrow x^4+2ax^3+\left(a^2-b^2-c^2)x^2+ab^2=0} ,

    and you get a rather nasty quartic equation which nevertheless can be solved with the help

    of Ferrari-Cardano formulae (pretty frightening creatures!), or perhaps you know some relation

    between the coefficients a,b,c which could simplify the above (for example, a=0 could make things waaaaaay simpler)

    Tonio
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  6. #6
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    Rearrange in terms of x?-pic.jpgClick image for larger version. 

Name:	Pic.jpg 
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ID:	20540No, definitely out of my depth. May have to try and find a simpler method. Was trying to find an equation to determine x on the attached diagram given a,b and c. Came up with the above. Is there a simpler approach to this?

    Thanks again.
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  7. #7
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    Quote Originally Posted by EdCoventry View Post
    Click image for larger version. 

Name:	Pic.jpg 
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ID:	20540Click image for larger version. 

Name:	Pic.jpg 
Views:	14 
Size:	11.3 KB 
ID:	20540No, definitely out of my depth. May have to try and find a simpler method. Was trying to find an equation to determine x on the attached diagram given a,b and c. Came up with the above. Is there a simpler approach to this?

    Thanks again.

    Sorry, what I wrote here's wrong since we don't know what the horizontal leg's length is.

    Tonio
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  8. #8
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    Tonio,

    Oops, sorry about that. New to this. Should it be a in (a + b)^2 though? I thought that would be different term?

    Apologies, please ignore, just seen your reply.
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  9. #9
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    Quote Originally Posted by EdCoventry View Post
    Tonio,

    Oops, sorry about that. New to this. Should it be a in (a + b)^2 though? I thought that would be different term?

    Apologies, please ignore, just seen your reply.

    You sent the sketch of a straight-angle triangle with hypotenuse equal to  a+x , vertical leg equal to c and

    horizontal leg equal to b+? . It's the "?" part that we don't know what prevents from us to use Pythagoras Theorem

    Tonio
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  10. #10
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    Thanks Tonio. Looks like there's no simple solution then. Am left with trying to solve the nasty quartic equation... I think beyond the limit of my abilities
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  11. #11
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    Hi,

    Quote Originally Posted by e^(i*pi) View Post
    x^2 = b^2 + \left(c - \dfrac{ac}{a+x}\right)^2

    A good place to start would be to use foil to expand that squared term

    \left(c - \dfrac{ac}{a+x}\right)^2 = c^2 - \dfrac{2ac}{a+x} + \dfrac{a^2c^2}{(a+x)^2}

    You should then be able to multiply through by (a+x)^2 to clear the fraction

    x^2(a+x)^2 = b^2(a+x)^2 + c^2(a+x)^2 - 2ac(a+x) + a^2c^2


    You will eventually have to expand it and it won't be pretty but it should be simple enough using FOIL and the distributive property
    Having another go at this. Should that second equation have - 2ac^2 / (a + x) rather than - 2ac / a+x?

    Thanks again
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  12. #12
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    Quote Originally Posted by EdCoventry View Post
    Hi,



    Having another go at this. Should that second equation have - 2ac^2 / (a + x) rather than - 2ac / a+x?

    Thanks again
    Yes it should. I made an expansion error (which is why you don't try complex problems after midnight)

    x^2(a+x)^2 = b^2(a+x)^2 + c^2(a+x)^2 - 2ac^2(a+x) + a^2c^2

    x^2(a+x)^2 = x^2(a^2+2ax+x^2) = a^2x^2+2ax^3+x^4

    b^2(a+x)^2 = b^2(a^2+2ax+x^2) = a^2b^2+2ab^2x+x^2b^2

    c^2(a+x)^2 = c^2(a^2+2ax+x^2) = a^2c^2+2ac^2x+c^2x^2

    2ac^2(a+x) = 2a^2c^2+2ac^2x

    -----------------------------

    a^2x^2+2ax^3+x^4 = (a^2b^2+2ab^2x+x^2b^2)+ (a^2c^2+2ac^2x+c^2x^2) - (2a^2c^2+2ac^2x) + a^2c^2

    Collect those with an x term on the right hand side.

    [(2ab^2x +x^2b^2) + (2ac^2x+c^2x^2) - 2ac^2x] + (a^2b^2+a^2c^2-2a^2c^2+a^2c^2)

    Cancelling gives

    2ab^2x +x^2b^2 + c^2x^2 + a^2b^2

    This is equal to the LHS

    a^2x^2+2ax^3+x^4 = 2ab^2x +x^2b^2 + c^2x^2 + a^2b^2


    x^4+2ax^3+a^2x^2-2ab^2x-x^2b^2-c^2x^2 - a^2b^2 = 0

    Collect like terms

    x^4 + 2ax^3 + (a^2-b^2-c^2)x^2 - 2ab^2x - a^2b^2 = 0

    Which is a nasty quartic equation but will probably be easier if you use an iterative method
    Last edited by e^(i*pi); January 22nd 2011 at 03:50 AM. Reason: dodgy signs
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  13. #13
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    Excellent, thanks for that. When I expand that I get:

    x^4 + 2ax^3 + (a^2 - b^2 - c^2)x^2 - (2ab^2)x - (a^2)(b^2) = 0

    Is this right as it's different to Tonio's expansion (also apologies for the format, haven't worked that out yet).

    Thanks again.
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  14. #14
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    e^(i*pi)'s Avatar
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    I edited working into my above post, it looks right but it would be only too easy to make a mistake but at least we get the same answer!
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  15. #15
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    Brilliant. That's what I got. The only difference being the + 2ab^2x in the last step. Should that be - 2ab^2x.
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