# Algebriac relations

• Oct 26th 2009, 02:36 PM
scrible
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(Worried)
• Oct 26th 2009, 09:46 PM
ramiee2010
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)

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

(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 $\displaystyle \alpha^3 - p \alpha^2 \ and \quad \beta^3 - p \beta^2$
$\displaystyle x^2-( sum\ of \ roots)x +(product\ of\ roots)=0$
$\displaystyle x^2-(-pq)x+q^3=0$
$\displaystyle x^2+pqx+q^3=0$
• Oct 26th 2009, 09:57 PM
ramiee2010
(5)
Given that the roots of the equation http://www.mathhelpforum.com/math-he...810a5f56-1.gif are http://www.mathhelpforum.com/math-he...ed813421-1.gif and http://www.mathhelpforum.com/math-he...0f951104-1.gif,

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 ,