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 August 4th, 2013, 09:22 PM #1 Member   Joined: Jul 2013 Posts: 58 Thanks: 0 Alpha/Beta and Beta/Alpha (quadratic roots) If Alpha and Beta be the roots of the equation 2x^2-6x+5=0 , find the equation whose roots are: $\alpha/\beta , \beta/\alpha$ I know they both are symmetric. a= 2 , b=-6 and c=5. alpha+beta = -6/2 and alpha*beta = 5/2. How to use them? I am confused again.. I am not getting how to get the next step. Besides I don't have proper book... the text book I have is full of errors and they call it a self-teaching guide. :/
August 4th, 2013, 10:39 PM   #2
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Re: Alpha/Beta and Beta/Alpha (quadratic roots)

Quote:
 Originally Posted by Dean1884 If Alpha and Beta be the roots of the equation 2x^2-6x+5=0 , find the equation whose roots are: $\alpha/\beta , \beta/\alpha$ I know they both are symmetric. a= 2 , b=-6 and c=5. alpha+beta = -6/2 and alpha*beta = 5/2. how to use them ? I am confused again.. I am not getting how to get the next step. Besides I don't have proper book.. the text book I have is full of errors and they call it a self-teaching guide. :/
Hello!
$\alpha + \beta=3$ ; $\frac {\alpha}{\beta}+1=\frac{3}{\beta}$ ; $\frac{\beta}{\alpha}+1=\frac{3}{\alpha}$ ; $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2=\frac{ 3(\beta+\alpha)}{\alpha \beta}=\frac{18}{5}$ ; $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{8} {5}$ and $\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}=1$........
Thank You!

 August 5th, 2013, 02:30 AM #3 Member   Joined: Jul 2013 Posts: 58 Thanks: 0 Re: Alpha/Beta and Beta/Alpha (quadratic roots) Thanks Dacu. The book says the answer is 5x^2-8x+5 = 0. How to turn $\alpha+\beta$ and $\alpha*\beta$ into this form $\alpha/\beta , \beta/\alpha$ ? What should I do in such situations?
August 5th, 2013, 03:59 AM   #4
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Re: Alpha/Beta and Beta/Alpha (quadratic roots)

Quote:
 Originally Posted by Dean1884 Thanks Dacu. The book says the answer is 5x^2-8x+5 = 0. How to turn $\alpha+\beta$ and $\alpha*\beta$ into this form $\alpha/\beta , \beta/\alpha$ ? What should I do in such situations?
$x^2-S\cdot x+P=0$ ; $S=x_1+x_2=\frac{\alpha}{\beta}+\frac{\beta}{\alpha }=\frac{8}{5}$ ; $P=x_1 \cdot x_2=\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}=1$ and so the equation is $x^2\,-\,\frac{8}{5} \cdot x\,+\,1\,=\,0$ and it is equivalent to the equation $5x^2\,-\,8x\,+\,5\,=\,0$.
Thank you!

August 5th, 2013, 08:03 AM   #5
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Quote:
 Originally Posted by Dean1884 alpha+beta = -6/2 and alpha*beta = 5/2. How to use them? How to turn $\alpha+\beta$ and $\alpha*\beta$ into this form $\alpha/\beta,\, \beta/\alpha$?
$\frac\alpha\beta\,+\,\frac\beta\alpha\,=\,\frac{\a lpha^2 + \beta^2}{\alpha\beta}\,=\,\frac{(\alpha\,+\,\beta) ^2}{\alpha\beta}\,-\,2\,=\,\frac{3^2}{5/2}\,-\,2\,=\,\frac85\text{ and }\frac\alpha\beta\cdot\frac\beta\alpha\,=\,1,\text { etc.}$

 August 5th, 2013, 12:16 PM #6 Member   Joined: Jul 2013 Posts: 58 Thanks: 0 Re: Alpha/Beta and Beta/Alpha (quadratic roots) oh ok. thanks.
 August 5th, 2013, 12:35 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 If copying from your initial attempt, note that $^{\alpha\,+\,\beta}$ is 6/2, not -6/2.
 August 5th, 2013, 12:45 PM #8 Member   Joined: Jul 2013 Posts: 58 Thanks: 0 Re: Alpha/Beta and Beta/Alpha (quadratic roots) all right.

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