# Thread: The roots of the equation

1. ## The roots of the equation

There a another problem that is hard to figure. The question ask that the roots of the equation $\displaystyle 9x^2+6x+1 = 4kx$ where $\displaystyle k$ is a real constant, are denoted by $\displaystyle \alpha$ and $\displaystyle \beta$.
(a) Show that the equation whose roots are $\displaystyle \frac {1}{\alpha}$ and $\displaystyle \frac {1}{\beta}$ is $\displaystyle x^2+6x+9=4kx$
(b) find the set of values of $\displaystyle k$ for which $\displaystyle \alpha$ and $\displaystyle \beta$ are real and positive. Can anyone help me with this problem?

2. Originally Posted by scrible
There a another problem that is hard to figure. The question ask that the roots of the equation $\displaystyle 9x^2+6x+1 = 4kx$ where $\displaystyle k$ is a real constant, are denoted by $\displaystyle \alpha$ and $\displaystyle \beta$.
(a) Show that the equation whose roots are $\displaystyle \frac {1}{\alpha}$ and $\displaystyle \frac {1}{\beta}$ is $\displaystyle x^2+6x+9=4kx$
(b) find the set of values of $\displaystyle k$ for which $\displaystyle \alpha$ and $\displaystyle \beta$ are real and positive. Can anyone help me with this problem?
HI

$\displaystyle 9x^2+6x+1-4kx=0$

$\displaystyle x^2+[\frac{6-4k}{9}]x+\frac{1}{9}=0$

$\displaystyle \alpha+\beta=\frac{4k-6}{9}$

$\displaystyle \alpha\beta=\frac{1}{9}$

$\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\bet a}{\alpha\beta}$

$\displaystyle =4k-6$

$\displaystyle \frac{1}{\alpha}(\frac{1}{\beta})=9$

the new equation would be $\displaystyle x^2-(4k-6)x+9=0$

then expand it

for (b) you should realise that if the roots are real , then

$\displaystyle b^2-4ac\geq 0$

3. Originally Posted by scrible
There a another problem that is hard to figure. The question ask that the roots of the equation $\displaystyle 9x^2+6x+1 = 4kx$ where $\displaystyle k$ is a real constant, are denoted by $\displaystyle \alpha$ and $\displaystyle \beta$.
(a) Show that the equation whose roots are $\displaystyle \frac {1}{\alpha}$ and $\displaystyle \frac {1}{\beta}$ is $\displaystyle x^2+6x+9=4kx$
(b) find the set of values of $\displaystyle k$ for which $\displaystyle \alpha$ and $\displaystyle \beta$ are real and positive. Can anyone help me with this problem?
$\displaystyle 9x^2+6x+1 = 4kx$
$\displaystyle 9x^2+6x-4kx+1 =0$
$\displaystyle 9x^2+(6-4k)x+1 =0$
if $\displaystyle ax^2+bx+c=0$
find a, b and c from the equation...
now, you have to determine sum and product root from the equation.
$\displaystyle \alpha$+$\displaystyle \beta$=\frac{-b}{a}=....
and
$\displaystyle \alpha$ x $\displaystyle \beta$=\frac{c}{a}=....

a) construct the equation whose roots are $\displaystyle \frac {1}{\alpha}$ and $\displaystyle \frac {1}{\beta}$
first you have to find:
- sum roots: $\displaystyle \frac {1}{\alpha}$+[tex]\frac {1}{\beta}=...
- product roots:$\displaystyle \frac {1}{\alpha}$[tex]\frac {1}{\beta}=...

after you determine the sum and product, the equation would be:
$\displaystyle x^2-(sum)x+(product)=0$

....hope it help....

HI

$\displaystyle 9x^2+6x+1-4kx=0$

$\displaystyle x^2+[\frac{6-4k}{9}]x+\frac{1}{9}=0$

$\displaystyle \alpha+\beta=\frac{4k-6}{9}$

$\displaystyle \alpha\beta=\frac{1}{9}$

$\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\bet a}{\alpha\beta}$

$\displaystyle =4k-6$

$\displaystyle \frac{1}{\alpha}(\frac{1}{\beta})=9$

the new equation would be $\displaystyle x^2-(4k-6)x+9=0$

then expand it

for (b) you should realise that if the roots are real , then

$\displaystyle b^2-4ac\geq 0$
Thank you very much. You were of great help. I am greatful.