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Evaluate 2/α + 3β².

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anemone

MHB POTW Director
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Feb 14, 2012
3,719
Problem:

Given \(\displaystyle \alpha\) and \(\displaystyle \beta\) are roots of the equation \(\displaystyle x^2-7x+8=0\), where \(\displaystyle \alpha>\beta\). Find the value of \(\displaystyle \frac{2}{\alpha}+3\beta^2\) without solving the equation.

Hi members of the forum, I just couldn't remember where did I find this problem but as I have tried to solve for the value of the intended expression, I kind of believe this couldn't be done without actually having to solve the given equation. But I wasn't sure.

Could someone please show me how to find the value of \(\displaystyle \frac{2}{\alpha}+3\beta^2\) without solving the equation, please?

Thanks.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: Evaluate 2/α+3β².

By solving the equation you mean finding explcitly the values of \(\displaystyle \alpha \, , \, \beta \) ?
 

mathbalarka

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MHB Math Helper
Mar 22, 2013
573
Re: Evaluate 2/α+3β².

I don't know if there is any other way to do it, but I did it this way :

Let's call \(\displaystyle T = T(\alpha, \beta) = \frac{2}{\alpha} + 3 \beta^2\).

The first thing comes into mind is whether this is symmetric or not. It's certainly not symmetric, but we can set another expression which permutes the position of the roots in T :

\(\displaystyle T' = T(\beta, \alpha) = \frac{2}{\beta} + 3 \alpha^2\)

Summing up T and T', we get :

\(\displaystyle T + T' = 2 \left ( \frac{1}\alpha + \frac{1}{\beta} \right ) + 3 ( \alpha^2 + \beta^2 )\)

\(\displaystyle \Rightarrow T + T' = \frac{\alpha + \beta}{\alpha \beta} + 3 s_2\)

Where s2 is one of the newton's symmetric polynomial. Hence, evaluating the expression, we get :

\(\displaystyle T + T' = \frac{415}{4}\)

Similarly, by multiplication,

\(\displaystyle T \cdot T' = \frac{2473}{4}\)

Now this can be easily solved. I recommend you to check my numerical calculations to see if I am correct. I am not giving the final value for T since it's very trivial to solve now. However, I like this problem. it's real toughie.
 
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anemone

MHB POTW Director
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Feb 14, 2012
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Re: Evaluate 2/α+3β².

By solving the equation you mean finding explcitly the values of \(\displaystyle \alpha \, , \, \beta \) ?
Yes, Zaid!:)
 

soroban

Well-known member
Feb 2, 2012
409
Re: Evaluate 2/α+3β².

Hello, anemone!

I can get only so far ... then I'm stuck, too.


Given \(\displaystyle \alpha\) and \(\displaystyle \beta\) are roots of the equation \(\displaystyle x^2-7x+8=0\), where \(\displaystyle \alpha>\beta\).
Find the value of \(\displaystyle X \,=\,\tfrac{2}{\alpha}+3\beta^2\) without solving the equation.

We have: .[tex]\begin{Bmatrix}\alpha + \beta &=& 7 & [1] \\ \alpha\beta &=& 8 & [2] \end{Bmatrix}[/tex]

Multiply [1] and [2]:

. . [tex]\alpha\beta(\alpha + \beta) \:=\:8\cdot7 \quad\Rightarrow\quad \alpha^2\beta + \alpha\beta^2 \:=\:56 \quad\Rightarrow\quad \alpha\beta^2 \:=\:56 - \alpha^2\beta\;\;[3][/tex]


We are given: .[tex]X \;=\;\frac{2}{\alpha} + 3\beta^2 \;=\;\frac{2+3\alpha\beta^2}{\alpha}[/tex]

Substitute [3]: .[tex]X \;=\;\frac{2+3(56-\alpha^2\beta)}{\alpha} \;=\;\frac{2+168-3\alpha^2\beta}{\alpha} [/tex]

. . . . . [tex]X \;=\; \frac{170 - 3\alpha^2\beta}{\alpha} \;=\;\frac{170}{\alpha} - \frac{3\alpha^2\beta}{\alpha} \;=\;\frac{170}{\alpha} - 3\alpha\beta[/tex]

Substitute [2]: .[tex]X \;=\;\frac{170}{\alpha} - 3(8) \;=\;\frac{170}{\alpha} - 24[/tex]


I see no way to evaluate [tex]X[/tex] without the value of [tex]\alpha.[/tex]