The general form of a quadratic equation is \(ax^2+bx+c=0\), where \(a,b\) and \(c\) are constants and \(a\neq 0\). Dividing by \(a\) throughout, we get

\[x^2+\frac{b}{a}x+\frac{c}{a}=0\cdots(1)\]

If the roots of the equation are \(\alpha\) and \(\beta\), we can then write the equation with roots \(\alpha\) and \(\beta\) in the form

\[x=\alpha \quad \text{or} \quad x=\beta\]

\[\implies (x-\alpha)(x-\beta)=0\]

\[\implies x^2-(\alpha+\beta)x+\alpha\beta=0\cdots(2)\]

Comparing \((1)\) and \((2)\), we see that

\[\text{Sum of roots}=\alpha+\beta=-\frac{b}{a}\]

\[\text{Product of roots}=\alpha\beta=\frac{c}{a}\]

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TopNewestNice set! The problems are hard but managed to solve them all. – Daniel Lim · 2 years, 3 months ago

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Great note @Victor Loh – Mardokay Mosazghi · 2 years, 3 months ago

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Thank you :) @Mardokay Mosazghi – Victor Loh · 2 years, 3 months ago

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