Simple roots of a quadratic question

1. Nov 18, 2007

rock.freak667

1. The problem statement, all variables and given/known data
Given that the roots of $x^2+px+q=0$ are $\alpha and \beta$, form an equation whose roots are $\frac{1}{\alpha} and \frac{1}{\beta}$
b) Given that $\alpha$ is a root of the equation $x^2=2x-3$ show that
i)$\alpha^3=\alpha-6$
ii)$\alpha^2-2\alpha^3=9$

2. Relevant equations
$$\sum\alpha=\frac{-b}{a}$$
$$\sum\alpha\beta=\frac{c}{a}$$

3. The attempt at a solution
well for part a) I can do that simply
but it is part b) that confuses me, as I can simply factorize what the eq'n and get 1 and -3 as the roots..but I don't see how i can use those numbers to prove what they want to show...

Last edited: Nov 18, 2007
2. Nov 18, 2007

dynamicsolo

Are you quite sure the roots are 1 and -3? What is the discriminant?

3. Nov 18, 2007

rock.freak667

ah wait...wrong sign i used...must redo

Well the roots are imaginary, but if i take $\alpha$ to be one of them and cube it I wouldn't get what i want

Last edited: Nov 18, 2007
4. Nov 18, 2007

dynamicsolo

What exactly were the roots you found? I just tried out (b-i) and got that to work. I'm trying (b-ii) now...

5. Nov 18, 2007

rock.freak667

1+- 2sqrt(2)i

6. Nov 18, 2007

dynamicsolo

I find the discriminant to be -8 , so the imaginary part would be just sqrt(2).

7. Nov 18, 2007

rock.freak667

ah...forgot to divide the imag. part by 2...but if i take $\alpha$ as either of the 2 roots it will work out?

8. Nov 18, 2007

dynamicsolo

You'll find that changing the sign of the imaginary part only flips the signs of certain terms in the expressions you are evaluating in a manner that still allows them to cancel out. Try it on (b-i)...

9. Nov 18, 2007

HallsofIvy

Staff Emeritus
Of course, actually solving the equation and using those values as $\alpha$ defeats the point of the problem: use the equation itself!

If $\alpha$ satisfies $x^2= 2x- 3$ then obviously $\alpha^2= 2\alpha - 3$ so $\alpha^3= 2\alpha^2- 3\alpha$. But, again, $\alpha^2= 2\alpha - 3$ so $\alpha^3= 2(2\alpha- 3)- 3\alpha= \alpha - 6$