# Sum and Product of Roots - Theory

Suppose that the quadratic equation

\[a{x^2} + bx + c = 0,\,\,\,a \ne 0\]

has the roots \(\alpha \) and \(\beta \). Thus, by the factor theorem, \(\left( {x - \alpha } \right)\) and \(\left( {x - \beta } \right)\) will be factors of the expression \(a{x^2} + bx + c\):

\[a{x^2} + bx + c = a\left( {x - \alpha } \right)\left( {x - \beta } \right)\]

A factor of \(a\) is required to equalize the coefficients on both sides. If we expand the right hand side, we have:

\[\begin{align}&a{x^2} + bx + c = a\left( {{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta } \right)\\&\Rightarrow \,\,\,a{x^2} + bx + c = a{x^2} - a\left( {\alpha\,+\, \beta } \right)x + a\alpha \beta \end{align}\]

By comparing the coefficients on both sides, we obtain expressions for the sum and product of the roots \(\alpha \) and \(\beta \):

\[\begin{align}&- a\left( {\alpha\, +\, \beta } \right) = b\,\,\, \Rightarrow \,\,\,S = \alpha\, + \,\beta\, =\, - \frac{b}{a}\\&a\,\alpha \beta\, =\, c\,\,\, \Rightarrow \,\,\,P = \alpha \beta\, = \,\frac{c}{a}\end{align}\]

Given a quadratic equation, we can evaluate the sum and product of its roots using these expressions. Here are two examples (we will use \(\alpha \) and \(\beta \) to denote the two roots):

\[\begin{align}&{x^2} - 3x + 2 = 0\\&\Rightarrow \,\,\,\left\{ \begin{array}{l}\alpha + \beta = - \frac{b}{a} = - \frac{{\left( { - 3} \right)}}{1} = 3\\\alpha \beta = \frac{c}{a} = \frac{2}{1} = 2\end{array} \right.\\&2{x^2} - 5x + 2 = 0\\&\Rightarrow \,\,\,\left\{ \begin{array}{l}\alpha + \beta = - \frac{b}{a} = - \frac{{\left( { - 5} \right)}}{2} = \frac{5}{2}\\\alpha \beta = \frac{c}{a} = \frac{2}{2} = 1\end{array} \right.\end{align}\]

We can also do the reverse: given the sum and product of the roots of a quadratic equation, we can obtain the equation. If the roots are \(\alpha \) and \(\beta \), we can write the equation as:

\[\begin{align}&\left( {x - \alpha } \right)\left( {x - \beta } \right) = 0\\&\Rightarrow \,\,\,{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta = 0\\&\Rightarrow \,\,\,{x^2} - Sx + P = 0\end{align}\]

For example, suppose that a quadratic equation is such that \(S = 5\) and \(P = 4\). This quadratic equation will be:

\[\begin{align}&{x^2} - Sx + P = 0\\&\Rightarrow \,\,\,{x^2} - 5x + 4 = 0\end{align}\]

As another example, suppose that \(S = \frac{3}{2}\) and \(P = - \frac{1}{2}\) for a quadratic equation. This equation will be:

\[\begin{align}&{x^2} - Sx + P = 0\\&\Rightarrow \,\,\,{x^2} - \frac{3}{2}x - \frac{1}{2} = 0\\&\Rightarrow \,\,\,2{x^2} - 3x - 1 = 0\end{align}\]