Symmetric Functions of the Roots
Symmetric Functions of the Roots
Exam questions rarely ask for \(\alpha\) and \(\beta\) themselves. They ask for combinations such as \(\alpha^2+\beta^2\) or \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\). An expression is symmetric when swapping \(\alpha\) and \(\beta\) leaves it unchanged — and every symmetric expression can be evaluated from just \(\alpha+\beta\) and \(\alpha\beta\), without solving the equation.
Key identities
| Expression | In terms of \(\alpha+\beta\) and \(\alpha\beta\) |
|---|---|
| \(\alpha^2+\beta^2\) | \((\alpha+\beta)^2-2\alpha\beta\) |
| \((\alpha-\beta)^2\) | \((\alpha+\beta)^2-4\alpha\beta\) |
| \(\alpha^3+\beta^3\) | \((\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)\) |
| \(\alpha^3-\beta^3\) | \((\alpha-\beta)\left[(\alpha+\beta)^2-\alpha\beta\right]\) |
| \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\) | \(\dfrac{\alpha+\beta}{\alpha\beta}\) |
| \(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\) | \(\dfrac{\alpha^2+\beta^2}{\alpha\beta}\) |
\(\alpha-\beta\) itself is not symmetric — swapping the roots changes its sign. Find \((\alpha-\beta)^2\) first, then take the square root and use the given ordering (such as \(\alpha>\beta\)) to choose the sign.
Do not memorise every variant. Start from \((\alpha+\beta)^2\) or \((\alpha+\beta)^3\), expand, and rearrange to isolate the expression you need.
The reciprocal-root identities and the displayed ratios require both roots to be nonzero, so \(\alpha\beta\ne0\). Check this condition before using those identities or forming an equation with reciprocal roots.
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