# Sum and Product of Roots - Theory

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}

grade 10 | Questions Set 2