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Math Help - Simulatenous Quadratic Equations

  1. #1
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    Simulatenous Quadratic Equations

    Can anybody solve these two equations with two unknowns for me?

    \frac{a}{a+b}= \frac{4}{102}, and

    \frac{a*b}{(a+b)^2(a+b+1)} = 0.0003658


    Basically I am doing a Statistics question about beta distribution and this has cropped up during the question, and when i get my answers, i keep gettin 'a' as a negative number (which is not permitted in beta distribution)

    Thanks
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  2. #2
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    Quote Originally Posted by sirellwood View Post
    Can anybody solve these two equations with two unknowns for me?

    \frac{a}{a+b}= \frac{4}{102}, and

    \frac{a*b}{(a+b)^2(a+b+1)} = 0.0003658


    Basically I am doing a Statistics question about beta distribution and this has cropped up during the question, and when i get my answers, i keep gettin 'a' as a negative number (which is not permitted in beta distribution)

    Thanks
    From \frac{a}{a+ b}= \frac{4}{102}= \frac{2}{51}
    51a= 2(a+ b)= 2a+ 2b
    49a= 2b
    b= \frac{49}{2}a

    Replace the "b" in \frac{ab}{(a+ b)(a+ b+ 1}= 0.0003658
    with that and you will have a cubic equation for a.
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  3. #3
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    thanks, yeah i got a=4 and b=98 (3 s.f) better answers!
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  4. #4
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    Hello, sirellwood!

    Can anybody solve these two equations with two unknowns for me?

    . . \dfrac{a}{a+b}\:=\:\dfrac{4}{102} \qquad \dfrac{ab}{(a+b)^2(a+b+1)} \:=\: 0.0003658


    Basically I am doing a Statistics question about beta distribution
    and this has cropped up during the question.
    And when i get my answers, i keep gettin 'a' as a negative number. ??

    First, I'll re-write the equations:

    . . [1] . \dfrac{a}{a+b} \:=\:\dfrac{2}{51}

    . . [2] . \dfrac{ab}{(a+b)^2(a+b+1)} \:=\:d\;\text{ (for decimal)}


    From [1], we have: . \dfrac{a}{a+b} \:=\:\dfrac{2}{51} \quad\Rightarrow\quad b \:=\:\dfrac{49}{2}a .[3]


    From [2], we have:

    . . \displaystyle \underbrace{\left(\frac{a}{a+b}\right)}_{\text{Thi  s is }\frac{2}{51}}\cdot \frac{b}{(a+b)(a+b+1)} \:=\:d

    . . . . . . \displaystyle \frac{2}{51}\cdot\frac{b}{(a+b)(a+b+1)} \:=\:d


    . . . . . . . . . . . . . . . . . . . . 2b \;=\;51d(a+b)(a+b+1)


    Substitute [3]:

    . . 2\left(\frac{49}{2}a\right) \;=\;51d\left(a + \frac{49}{2}a\right)\left(a + \frac{49}{2}a + 1\right)

    . . . . . 49a \;=\;51d\left(\frac{51}{2}a\right)\left(\frac{51a+  2}{2}\right)


    \text{Divide by }a\!:\;\;49 \:=\:\dfrac{2601d(51a+2)}{4} \quad\Rightarrow\quad 196 \;=\;132,\!651ad + 5202d

    . . . . . 132,\!651ad \;=\;196 - 5202d \quad\Rightarrow\quad a \;=\;\dfrac{196-5202d}{132,\!651d}


    Subsstitute d = 0.0003658

    . . a \;=\;\dfrac{196 - 5202(0.0003658)}{132,\!651(0.0003658)} \;=\;4.000044621


    Therefore: . a\:\approx\;4



    Edit: Too slow again . . . *sigh
    .
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