Creating a Quadratic Equation with Inverted Roots

[Maatttt0]

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 :)

 

[LCKurtz]

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?

 

[Maatttt0]

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 :)

 

[Hurkyl]

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?

 

[Maatttt0]

Do you mean;

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

$$x² - \alpha x - \beta x + \alpha\bet = 0$$

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

 

[LCKurtz]

Do you mean;

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

$$x² - \alpha x - \beta x + \alpha\bet = 0$$

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?

 

[Maatttt0]

Yes it is - okay thank you, I shall give it ago :)