# Construct equation.

• Dec 17th 2009, 04:52 AM
Detanon
Construct equation.
If α, β are the roots of the equation x^2 - 5x + 3 = 0, construct an equation with roots α/ β^2 and β/ α^2.

Help I'm having trouble with this q,i got -18/25 for the sum and 1/5 for product. But the answer is wrong. I think im calculating the product wrong when i multiply.
Can someone show the steps to do this please. Thanks!
• Dec 17th 2009, 05:48 AM
dedust
Quote:

Originally Posted by Detanon
If α, β are the roots of the equation x^2 - 5x + 3 = 0, construct an equation with roots α/ β^2 and β/ α^2.

Help I'm having trouble with this q,i got -18/25 for the sum and 1/5 for product. But the answer is wrong. I think im calculating the product wrong when i multiply.
Can someone show the steps to do this please. Thanks!

let $\displaystyle \mu = \frac{\alpha}{\beta^2}$ and $\displaystyle \gamma = \frac{\beta}{\alpha^2}$ be the roots of the new polynomial, then

$\displaystyle \mu + \gamma = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2}$$\displaystyle =\frac{\alpha^3 + \beta^3}{(\alpha \beta)^2} = \frac{(\alpha + \beta)^3 - 3\alpha \beta(\alpha + \beta)}{(\alpha \beta)^2}$

and

$\displaystyle \mu \gamma = \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} = \frac{1}{\alpha \beta}$

where
$\displaystyle \alpha + \beta = 5$ and $\displaystyle \alpha \beta = 3$

the new polynomial is $\displaystyle p(x) = x^2 - (\mu + \gamma)x + \mu \gamma$
• Dec 17th 2009, 07:52 AM
Soroban
Hello, Detanon!

I'll get you started . . .

Quote:

$\displaystyle \alpha, \beta$ are the roots of the equation $\displaystyle x^2 - 5x + 3 \:=\: 0$

Construct an equation with roots $\displaystyle \frac{\alpha}{\beta^2}$ and $\displaystyle \frac{\beta}{\alpha^2}$

The new equation will have:

. . x-coefficient: .$\displaystyle b \;=\;-\left(\frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2}\right) \;=\;-\frac{\alpha^3 + \beta^3}{\alpha^2\beta^2}$

. . Constant term: .$\displaystyle c \;=\;\left(\frac{\alpha}{\beta^2}\right)\left(\fra c{\beta}{\alpha^2}\right) \;=\;\frac{\beta}{\alpha}$

We have: .$\displaystyle \begin{array}{cccc}\alpha + \beta &=& 5 & [1] \\ \alpha\beta &=& 3 & [2] \end{array}$

Cube [1]: .$\displaystyle (\alpha + \beta)^3 \:=\:5^3 \quad\Rightarrow\quad \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3 \:=\:125$

. . $\displaystyle \alpha^3 + \beta^3 + 3\underbrace{\alpha\beta}_{3}\underbrace{(\alpha + \beta)}_{5} \;=\;125 \quad\Rightarrow\quad \alpha^3+\beta^3 \:=\:80$

Square [2]: .$\displaystyle (\alpha\beta)^2 \:=\:3^2 \quad\Rightarrow\quad \alpha^2\beta^2 \:=\:9$

Hence: .$\displaystyle b \;=\;-\frac{\alpha^3 + \beta^3}{\alpha^2\beta^2} \:=\:-\frac{80}{9}$

. . and we've found the x-coefficient . . .