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Math Help - Algebriac relations

  1. #1
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    Post Algebriac relations

    I did some promblems at the end of a math chapter, the last section is giving me some serious problem. Not shame to say that there is about five of them (2) Resolve the expression \frac {(x-2)}{(x^2+1)(x-1)^2} into its simplest partial fractions.
    (2) If \alpha and \beta are the roots of the equation ax^2 + bx +c = 0 show that the roots of the equation ax^2 - (b^2- 2ac)x + ac = 0 are \frac {\alpha}{\beta} and \frac {\beta}{\alpha}
    (3) The roots of the quadratic equation x^2-px+q=0 are \alpha and \beta Form, in terms of p and q, the quadratic equation whose roots are \alpha^3 - p\alpha^2 and \beta^3-p\beta^2.
    (4) Form a quadratic equation with roots which exceed by 2 and the roots of thequadratic equation 3x^2-(p-4)x-(2p+1)=0. Find the values of p for which the given equation has equal roots.
    (5) Given that the roots of the equation ax^2+bx+c=0 are \beta and n\beta, show that (n+1)^2ac=nb^2
    can some one or a few people ( as many as it can take) help me figure out these problem. I just want to understand the chapter thoroughly
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  2. #2
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    Please do not post more than 3 question in one post .

    Hints for the problem (1)

    partial fraction

    \frac {(x-2)}{(x^2+1)(x-1)^2} =\frac {Ax+B}{(x^2+1)}+\frac {C}{(x-1)}+\frac {D}{(x-1)^2}

    Hints for the problem (2),(3),(4) and (5)

    equation of quadratic equation
    \boxed { x^2-( sum\ of \ roots)x +(product\ of\ roots)=0 } \quad ......(1)

    (3) For x^2-px+q=0 given roots :  \alpha \ and\  \beta
     \therefore \ sum \ of \ roots=  \alpha + \beta= p \quad ....(2)
     \ product \ of \ roots= \alpha \cdot \beta= q \quad ...(3)

    the quadratic equation whose roots are \alpha^3 - p\alpha^2 and \beta^3 - q\beta^2
     sum\ of \ roots = (\alpha^3 - p\alpha^2)  + (\beta^3 - p\beta^2 )= \alpha^3  + \beta^3 -p (\alpha^2  + \beta^2)=-pq (\ how\ ? \ see\ spoiler)

    Spoiler:

      \alpha^3 - p\alpha^2 + \beta^3 - p\beta^2 =( \alpha^3 + \beta^3) -p (\alpha^2 + \beta^2)
     =(\alpha + \beta)(\alpha^2  + \beta^2-\alpha \beta)-p\{(\alpha  + \beta)^2 -2 \alpha \beta \}
    \color {blue} \{ \because a^3+b^3 =(a+b)(a^2-ab+b^2) \quad and \quad a^2+b^2 =(a+b)^2-2ab \}

     =(\alpha + \beta) \{(\alpha  + \beta)^2-2 \alpha \beta -\alpha \beta \} -p\{(\alpha  + \beta)^2 -2 \alpha \beta \}
     = (\alpha + \beta) \{(\alpha  + \beta)^2-3 \alpha \beta  \} -p\{(\alpha  + \beta)^2 -2 \alpha \beta \}
    substituting values of (\alpha  + \beta) \ and \ \alpha \beta from (2) and (3)
     = p \{ (p)^2-3q \} -p \{(p)^2-2q \}= p^3-3pq-p^3+2pq =-pq


    product of roots  = (\alpha^3 - p\alpha^2) \cdot (\beta^3 - p\beta^2 ) =p^3 \quad (how ? \ see \ spoiler)
    Spoiler:

     = (\alpha^3 - p\alpha^2) \cdot (\beta^3 - p\beta^2 )= \alpha^3 \beta^3 -p\alpha^3 \beta^2 -p \alpha^2 \beta^3 +p^2 \alpha^2 \beta^2
    =(\alpha \beta)^3 -p\alpha^2 \beta^2(\alpha+\beta) +p^2 ( \alpha \beta)^2)
     = (q)^3-p(q)^2 (p) +p^2(q)^2=q^3


    the quadratic equation whose roots are  \alpha^3 - p \alpha^2 \ and \quad \beta^3 - p \beta^2
     x^2-( sum\ of \ roots)x +(product\ of\ roots)=0
     x^2-(-pq)x+q^3=0
     x^2+pqx+q^3=0
    Last edited by ramiee2010; October 26th 2009 at 10:04 PM.
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  3. #3
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    (5)
    Given that the roots of the equation are and ,

    product of roots =  \beta \cdot n \beta =\frac{c}{a} \quad \Rightarrow  \beta ^2 =\frac{c}{a \cdot n} \quad .....(1)

    sum of roots =  \beta + n \beta= \frac{-b}{a} \quad \Rightarrow \beta + n \beta= \frac{-b}{a}
     \beta(1 + n)= \frac{-b}{a}
    squaring both side
      \beta ^2 (1 + n)^2= \frac{b^2}{a^2}
     \frac{c}{a \cdot n} (1 + n)^2= \frac{b^2}{a^2}
     \Rightarrow \frac{c}{a } (1 + n)^2 {a^2} = b^2 \cdot n
      \Rightarrow \boxed {  (1 + n)^2 ac =nb^2 }

    try (2) and (4) yourself ,
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