Creating a Quadratic Equation with Inverted Roots

AI Thread Summary
The discussion revolves around finding a quadratic equation with roots that are the inverses of the roots of the original equation x^2 + kx + 2k = 0, where k is a non-zero constant. Participants clarify that one root should indeed be inverted, leading to roots of the form α/β and β/α. The approach involves using the relationships between the roots and the coefficients of the quadratic equation, specifically αβ = 2k. There is also mention of a technical issue with the forum's preview feature affecting the discussion. Overall, the focus is on deriving a new quadratic equation based on the specified inverted roots.
Maatttt0
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Homework Statement



The quadratic equation x^2 + kx + 2k = 0 where k is a non-zero constant, has roots \alpha and \beta.

Find a quadratic equation with roots \frac{\alpha}{\beta} and \frac{\beta}{\alpha}. {one is meant to be inverted - the code isn't working properly :( }

Homework Equations





The Attempt at a Solution



A question like this, I would normal attempt by using 'u' as a variable and get \alpha on it's own, e.g:
\alpha + 1 = u
\alpha = u - 1

I would then subst. the u - 1 part into the equation and get the answer. I'm unsure where to go as alpha and beta are in a fraction together - help please :)
 
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Maatttt0 said:
Find a quadratic equation with roots \frac{\alpha}{\beta} and \frac{\alpha}{\beta}.

Just a question -- are those both intended to be the same or is one of them supposed to be inverted?
 
LCKurtz said:
Just a question -- are those both intended to be the same or is one of them supposed to be inverted?

Ahh thank you - meant to be inverted, I'll edit it now :)
 
I would have started by writing down the equation of the quadratic with those roots, and only then worry about trying to rewrite everything in terms of k.

What relationships do you know exist between alpha, beta, and k?
 
Do you mean;

(x - \alpha)(x - \beta)

x<sup>2</sup> - \alpha x - \beta x + \alpha\bet = 0

Therefore \alpha\beta = 2k...? :S
 
Maatttt0 said:
Do you mean;

(x - \alpha)(x - \beta)

x<sup>2</sup> - \alpha x - \beta x + \alpha\bet = 0

Therefore \alpha\beta = 2k...? :S

Don't use the sup in combination with tex tags. I think the equation you are trying to write is

x^2 -(\alpha + \beta)x + \alpha\beta = x^2 + kx + 2k

There is currently a bug in the "preview" feature for tex; the workaround is to hit the refresh F5 button on your browser after trying to preview.

Anyway, yes, that gives you a couple of equations to work with for α and β.

Note to Hurkyl: I have tried to get someone's attention about the preview bug in the HH forum and with a message to Greg. Do you know if anything is being looked at?
 
Yes it is - okay thank you, I shall give it ago :)
 
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