# Thread: Sum and Products of roots anyone?

1. ## Sum and Products of roots anyone?

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?

2. ## Re: Sum and Products of roots anyone?

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

3. ## Re: Sum and Products of roots anyone?

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

4. ## Re: Sum and Products of roots anyone?

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

5. ## Re: Sum and Products of roots anyone?

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).

6. ## Re: Sum and Products of roots anyone?

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

7. ## Re: Sum and Products of roots anyone?

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.

8. ## Re: Sum and Products of roots anyone?

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

9. ## Re: Sum and Products of roots anyone?

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?

10. ## Re: Sum and Products of roots anyone?

No way a 2U question.. haha.

11. ## Re: Sum and Products of roots anyone?

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...

12. ## Re: Sum and Products of roots anyone?

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.

13. ## Re: Sum and Products of roots anyone?

I got this from a tutor sheet

14. ## Re: Sum and Products of roots anyone?

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

15. ## Re: Sum and Products of roots anyone?

this resembles an extension question from cambridge ch8 on sum and product of roots, so it is def not 2u

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