MHB Evaluate 2/α + 3β²: Find Value w/o Equation Solving

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion centers on evaluating the expression 2/α + 3β², where α and β are roots of the quadratic equation x² - 7x + 8 = 0. Participants explore methods to find the value without explicitly solving for α and β. One approach involves using symmetric properties and Newton's symmetric polynomials, leading to expressions for T and T' that can be summed and multiplied. However, some members express difficulty in deriving a final value without knowing α explicitly, indicating that while techniques exist, they may not yield a straightforward solution. Ultimately, the consensus is that evaluating the expression without solving for the roots remains challenging.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Problem:

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

Thanks.
 
Mathematics news on Phys.org
Re: Evaluate 2/α+3β².

By solving the equation you mean finding explcitly the values of $$\alpha \, , \, \beta $$ ?
 
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 $$T = T(\alpha, \beta) = \frac{2}{\alpha} + 3 \beta^2$$.

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 :

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

Summing up T and T', we get :

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

$$\Rightarrow T + T' = \frac{\alpha + \beta}{\alpha \beta} + 3 s_2$$

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

$$T + T' = \frac{415}{4}$$

Similarly, by multiplication,

$$T \cdot T' = \frac{2473}{4}$$

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.
 
Last edited:
Re: Evaluate 2/α+3β².

ZaidAlyafey said:
By solving the equation you mean finding explcitly the values of $$\alpha \, , \, \beta $$ ?

Yes, Zaid!:)
 
Re: Evaluate 2/α+3β².

Hello, anemone!

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

Given $$\alpha$$ and $$\beta$$ are roots of the equation $$x^2-7x+8=0$$, where $$\alpha>\beta$$.
Find the value of $$X \,=\,\tfrac{2}{\alpha}+3\beta^2$$ without solving the equation.
We have: .\begin{Bmatrix}\alpha + \beta &=& 7 & [1] \\ \alpha\beta &=& 8 & [2] \end{Bmatrix}

Multiply [1] and [2]:

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

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

. . . . . X \;=\; \frac{170 - 3\alpha^2\beta}{\alpha} \;=\;\frac{170}{\alpha} - \frac{3\alpha^2\beta}{\alpha} \;=\;\frac{170}{\alpha} - 3\alpha\beta

Substitute [2]: .X \;=\;\frac{170}{\alpha} - 3(8) \;=\;\frac{170}{\alpha} - 24I see no way to evaluate X without the value of \alpha.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...

Similar threads

Replies
10
Views
1K
Replies
1
Views
2K
Replies
4
Views
2K
Replies
3
Views
1K
Replies
5
Views
2K
Replies
5
Views
2K
Back
Top