Relationship Between Roots and Coefficients of a Quadratic (Vieta's Theorem)

Consider the following quadratic:

Given that its roots are x = \alpha and x = \beta , the following can be deduced:

\begin{aligned}

x - \alpha &= 0 \

x - \beta &= 0

\end{aligned}

Therefore, the following equation can be used to describe the quadratic:

(x-\alpha)(x-\beta)=0

Which can be simplified to:

x^2-(\alpha+\beta)x+\alpha\beta=0

Call this equation Ⓐ.

This is the relationship that the roots and coefficients have with one another!

Now, recall the form of a general quadratic equation:

ax^2+bx+c=0,\quad \text{where } a,b,c\in\mathbb{R}

We can rearrange the above equation to allow the coefficient of x^2 to be 1, like in equation Ⓐ:

x^2+\frac{b}{a}x+\frac{c}{a}=0

Call this equation Ⓑ.

Now, compare the coefficients of the variables in both Ⓐ and Ⓑ:

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

\implies \alpha+\beta=-\frac{b}{a}

This is referred to as the ‘sum of roots’.

\alpha\beta=\frac{c}{a}

This is referred to as the ‘product of roots’.

Example 1: Find \alpha+\beta and \alpha\beta for -3x^2+6x-1=0.

Solution:

\begin{aligned}

a&=-3,\quad b=6,\quad c=-1\

\alpha+\beta&=-\frac{b}{a}=-\frac{6}{-3}=2\

\alpha\beta&=\frac{c}{a}=\frac{-1}{-3}=\frac{1}{3}

\end{aligned}

\therefore\quad \alpha+\beta=2,\quad \alpha\beta=\frac{1}{3}

Now, for exams, there are some common roots that can be asked of you to find. Note the following:

1.\quad \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta \[6pt]

2.\quad \alpha^2\beta^2=(\alpha\beta)^2 \[6pt]

3.\quad \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta} \[6pt]

4.\quad \frac{1}{\alpha\beta}=\frac{1}{\alpha\beta} \[6pt]

5.\quad \alpha-\beta=\sqrt{(\alpha+\beta)^2-4\alpha\beta} \[6pt]

6.\quad (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta