Simple roots of a quadratic question

[rock.freak667]

Homework Statement

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)$\alpha^2-2\alpha^3=9$

Homework Equations

$$\sum\alpha=\frac{-b}{a}$$
$$\sum\alpha\beta=\frac{c}{a}$$

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...

 

[dynamicsolo]

b) Given that $\alpha$ is a root of the equation $x^2=2x-3$ show that
i)$\alpha^3=\alpha-6$
ii)$\alpha^2-2\alpha^3=9$


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...

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

 

[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

 

[dynamicsolo]

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

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

 

[rock.freak667]

1+- 2sqrt(2)i

 

[dynamicsolo]

1+- 2sqrt(2)i

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

 

[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?

 

[dynamicsolo]

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

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)...

 

[HallsofIvy]

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 $\alpha^3= 2(2\alpha- 3)- 3\alpha= \alpha - 6$