Results 1 to 3 of 3

Thread: Algebriac relations

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    72

    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 $\displaystyle \frac {(x-2)}{(x^2+1)(x-1)^2}$ into its simplest partial fractions.
    (2) If $\displaystyle \alpha$ and $\displaystyle \beta$ are the roots of the equation $\displaystyle ax^2 + bx +c = 0$ show that the roots of the equation $\displaystyle ax^2 - (b^2- 2ac)x + ac = 0 $ are $\displaystyle \frac {\alpha}{\beta}$ and $\displaystyle \frac {\beta}{\alpha}$
    (3) The roots of the quadratic equation $\displaystyle x^2-px+q=0$ are $\displaystyle \alpha$ and $\displaystyle \beta $ Form, in terms of p and q, the quadratic equation whose roots are $\displaystyle \alpha^3 - p\alpha^2$ and $\displaystyle \beta^3-p\beta^2$.
    (4) Form a quadratic equation with roots which exceed by 2 and the roots of thequadratic equation $\displaystyle 3x^2-(p-4)x-(2p+1)=0$. Find the values of p for which the given equation has equal roots.
    (5) Given that the roots of the equation $\displaystyle ax^2+bx+c=0$ are $\displaystyle \beta$ and $\displaystyle n\beta$, show that $\displaystyle (n+1)^2ac=nb^2$
    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2009
    From
    New Delhi
    Posts
    153
    Please do not post more than 3 question in one post .

    Hints for the problem (1)

    partial fraction

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

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

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

    Spoiler:

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

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


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

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


    the quadratic equation whose roots are $\displaystyle \alpha^3 - p \alpha^2 \ and \quad \beta^3 - p \beta^2 $
    $\displaystyle x^2-( sum\ of \ roots)x +(product\ of\ roots)=0 $
    $\displaystyle x^2-(-pq)x+q^3=0 $
    $\displaystyle x^2+pqx+q^3=0 $
    Last edited by ramiee2010; Oct 26th 2009 at 10:04 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    From
    New Delhi
    Posts
    153
    (5)
    Given that the roots of the equation are and ,

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

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

    try (2) and (4) yourself ,
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relations and Functions - Inverse Relations Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Nov 13th 2011, 12:20 PM
  2. Replies: 1
    Last Post: Sep 19th 2011, 01:09 PM
  3. Relations in a set
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Sep 5th 2010, 10:03 PM
  4. Relations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 3rd 2009, 08:22 PM
  5. prove tht sqrt(\alpha) is algebriac over K
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Apr 10th 2008, 12:58 PM

Search Tags


/mathhelpforum @mathhelpforum