Sum and Products of Roots

Roots of Quadratic Equations

\begin{align}
& \left( x-\alpha \right)\left( x-\beta \right)=0 \\
& {{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0
\end{align}

\begin{align}
& a{{x}^{2}}+bx+c=0 \\
& {{x}^{2}}+\frac{b}{a}x+\frac{c}{a}=0 \\
\end{align}

Vieta’s Formulae:

Sum of Roots = \(\alpha +\beta \) = \(-\frac{b}{a}\)

Product of Roots = \(\alpha \beta =\frac{c}{a}\)

Things to note:

\begin{align}
& {{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab \\
& {{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}-ab+{{b}^{2}}) \\
& {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}}) \\
& {{a}^{4}}+{{b}^{4}}={{({{a}^{2}}+{{b}^{2}})}^{2}}-2{{a}^{2}}{{b}^{2}} \\
\end{align}

Using the sum and product of roots

1. Write the equation as \(ax^2+bx+c=0\) and read off \(\alpha+\beta=-\dfrac{b}{a}\) and \(\alpha\beta=\dfrac{c}{a}\).
2. Rewrite the required expression using only \(\alpha+\beta\) and \(\alpha\beta\).
3. To form an equation with new roots, find their sum \(S\) and product \(P\), then write \(x^2-Sx+P=0\).

Example

The roots of \(2x^2-6x+1=0\) are \(\alpha\) and \(\beta\). Find \(\alpha^2+\beta^2\) and an equation whose roots are \(\alpha^2\) and \(\beta^2\).

\(\alpha+\beta=3,\quad \alpha\beta=\dfrac{1}{2}\)

\(\begin{aligned} \alpha^2+\beta^2&=(\alpha+\beta)^2-2\alpha\beta \\ &=9-1 \\ &=8 \end{aligned}\)

\(\alpha^2\beta^2=(\alpha\beta)^2=\dfrac{1}{4}\)

\(\begin{aligned} x^2-8x+\dfrac{1}{4}&=0 \\ 4x^2-32x+1&=0 \end{aligned}\)