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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!

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    We can help you better if you show us how far you've gotten and what specifically is tripping you up. – 2017-01-11
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    Is there a reason you haven't just found the roots and evaluated? – 2017-01-11
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    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 – 2017-01-11

2 Answers 2

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Use Vieta's formula:

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?

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Hint:

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

Edit:

Hint 2:

Try to expand

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

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    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 :/ – 2017-01-11
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    hint $2$ might help. – 2017-01-11