Find sum of roots squared

Question

If $α$, $β$, $γ$, are the roots of the equation $x^3-5x+3=0$, find the value of $α^2+β^2+γ^2$.

Im stuck on this question for a while now, please help me with explanations, thank you very much!

We can help you better if you show us how far you've gotten and what specifically is tripping you up. — The Count, Jan 11 '17 at 20:54
That is one condition, we are not allowed to just find the roots and square them. We must use the product of roots and sum of roots rules, -b/a and c/a — 2000mroliver, Jan 11 '17 at 20:56

Siong Thye Goh · Answer 1 · 2017-01-11T21:03:52.750

5

Hint:

$$(\alpha^2+\beta^2+\gamma^2)=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\alpha\gamma)$$

Edit:

Hint 2:

Try to expand

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

edited Jan 11 '17 at 21:03

answered Jan 11 '17 at 20:55

Siong Thye Goh

thank you, that is actually as far as I had gotten but I could figure out how to write −2(αβ+βγ+αγ) in terms of the product or the sum :/ – 2000mroliver Jan 11 '17 at 21:02
hint $2$ might help. – Siong Thye Goh Jan 11 '17 at 21:04

score 1 · Accepted Answer · answered Jan 11 '17 at 21:03

Note that: $$(\alpha^2+\beta^2+\gamma^2)=(\alpha+\beta+\gamma)^2-2(\alpha \beta + \alpha \gamma + \beta \gamma)$$

Vieta's formula suggests for a polynomial $ax^3+bx^2+cx+d=0$:

$$(\alpha+\beta+\gamma)=-\frac{b}{a}$$

And:

$$(\alpha \beta + \alpha \gamma + \beta \gamma)=\frac{c}{a}$$

Can you take it from here?

2 Answers2