Sum and Products of roots anyone?

**Prawnchip** · 2 Jun 2012, 8:40 PM

If α and β are the roots of x² = 5x - 8, find ³√α + ³√β without finding the roots. Anyone know how to do this?

**Carrotsticks** · 2 Jun 2012, 10:41 PM

Form the polynomial with roots cube root alpha and cube root beta, then find the sum of roots using that polynomial.

**nightweaver066** · 2 Jun 2012, 10:44 PM

Originally Posted by Carrotsticks

Form the polynomial with roots cube root alpha and cube root beta, then find the sum of roots using that polynomial.

+1 much easier

**Sy123** · 2 Jun 2012, 11:01 PM

Originally Posted by Prawnchip

If α and β are the roots of x² = 5x - 8, find ³√α + ³√β without finding the roots. Anyone know how to do this?

Discriminant is less than zero
Moreover

Those are the LHS and RHS graphed. No intersection, therefore no roots
IMPOSSIBRU

**Carrotsticks** · 2 Jun 2012, 11:10 PM

It has no REAL roots, but doesn't mean no value of cube root alpha + cube root beta exists.

I think OP either mistyped the question, or posted in the wrong section (meant to be Extension 2).

**Sy123** · 2 Jun 2012, 11:15 PM

Heh, sorry about that then, I just...yeah

**deswa1** · 2 Jun 2012, 11:15 PM

Originally Posted by Timske

-7 < 0 , no roots
impossibruuu

This is true but you can still find sums and products for roots that aren't real. Consider the equation x^2+1=0. The discriminant is less than 0, therefore there are no real roots however the sums and products of roots gives us that the sum is 0 and the product is 1. When you do find the roots (using MX2 complex numbers), the roots are i and -i which do indeed add to zero and multiply to 1.

I think Carrotsticks is right though and this should be in the MX2 section.

**Timske** · 2 Jun 2012, 11:19 PM

Originally Posted by deswa1

This is true but you can still find sums and products for roots that aren't real. Consider the equation x^2+1=0. The discriminant is less than 0, therefore there are no real roots however the sums and products of roots gives us that the sum is 0 and the product is 1. When you do find the roots (using MX2 complex numbers), the roots are i and -i which do indeed add to zero and multiply to 1.

I think Carrotsticks is right though and this should be in the MX2 section.

oh i deleted that post lol, OP posted in wrong section then

**Prawnchip** · 2 Jun 2012, 11:54 PM

Sorry if it's in the wrong section i didn't know it was in mx2 :/ Carrotsicks can you explain? How do you form the polynomial?

**xdosha** · 3 Jun 2012, 12:31 AM

No way a 2U question.. haha.

**Fus Ro Dah** · 3 Jun 2012, 12:35 AM

Prawnchip, from where did you get this question?

For some reason, it's not working out nicely for me. The usual way would be to let $u= \sqrt[3]{\alpha}$ which implies that $\alpha = u^3$ then find $P(\alpha)=0$ but the polynomial transforms to a polynomial of degree six, rather than preserving the number of solutions...

**nightweaver066** · 3 Jun 2012, 1:00 AM

By using, $\alpha + \beta = (\alpha^{1/3} + \beta^{1/3})(\alpha^{2/3} - (\alpha \beta)^{1/3} + \beta^{2/3})$
and letting $x = \alpha^{1/3} + \beta^{1/3}$ , then rearranging to get:
x^3 - 6x - 5 = 0
(x + 1)(x^2 - x - 5) = 0

Not sure on which value i should take as the solution lol.

**Prawnchip** · 3 Jun 2012, 2:38 PM

I got this from a tutor sheet

**Nooblet94** · 3 Jun 2012, 4:51 PM

Originally Posted by nightweaver066

By using, $\alpha + \beta = (\alpha^{1/3} + \beta^{1/3})(\alpha^{2/3} - (\alpha \beta)^{1/3} + \beta^{2/3})$
and letting $x = \alpha^{1/3} + \beta^{1/3}$ , then rearranging to get:
x^3 - 6x - 5 = 0
(x + 1)(x^2 - x - 5) = 0

Not sure on which value i should take as the solution lol.

Pretty sure it's the positive solution, since $\alpha+\beta >0$

EDIT: Yep, it is, just checked on wolframalpha