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**\(\alpha-\beta\) 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.

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.

Why symmetric functions?

\(\alpha-\beta\) 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.