# The roots of the equation

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• Nov 15th 2009, 06:23 AM
scrible
The roots of the equation
There a another problem that is hard to figure. The question ask that the roots of the equation $9x^2+6x+1 = 4kx$ where $k$ is a real constant, are denoted by $\alpha$ and $\beta$.
(a) Show that the equation whose roots are $\frac {1}{\alpha}$ and $\frac {1}{\beta}$ is $x^2+6x+9=4kx$
(b) find the set of values of $k$ for which $\alpha$ and $\beta$ are real and positive. Can anyone help me with this problem?
• Nov 15th 2009, 06:34 AM
mathaddict
Quote:

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

HI

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

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

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

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

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

$=4k-6$

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

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

then expand it

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

$b^2-4ac\geq 0$
• Nov 15th 2009, 06:41 AM
pencil09
Quote:

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

$9x^2+6x+1 = 4kx$
$9x^2+6x-4kx+1 =0$
$9x^2+(6-4k)x+1 =0$
if $ax^2+bx+c=0$
find a, b and c from the equation...
now, you have to determine sum and product root from the equation.
$\alpha$+ $\beta$=\frac{-b}{a}=....
and
$\alpha$ x $\beta$=\frac{c}{a}=....

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

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

....hope it help....(Happy)
• Nov 15th 2009, 07:01 AM
scrible
Quote:

Originally Posted by mathaddict
HI

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

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

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

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

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

$=4k-6$

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

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

then expand it

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

$b^2-4ac\geq 0$

Thank you very much. You were of great help. I am greatful.