Relationships between the Roots and Coefficients of Quadratic/Tertiary/nth Degree Equations
Formulas
Relationship Between Roots and Coefficients of a Quadratic Equation
Let’s consider the roots of the quadratic equation $ax^{2}+bx+c=0$ to be $\alpha$ and $\beta$. Then the following relationship holds true.
$$ \alpha+\beta=-\frac{b}{a}\quad \& \quad \alpha\beta= \frac{ c}{a} $$
Relationship Between Roots and Coefficients of a Cubic Equation
Let’s consider the roots of the cubic equation $ax^{3}+bx^{2}+cx+d=0$ to be $\alpha$, $\beta$, and $\gamma$. Then the following relationship holds true.
$$ \alpha + \beta + \gamma = -\frac{b}{a}\quad \& \quad \alpha\beta + \beta \gamma +\gamma \alpha = \frac{c}{a}\quad \& \quad \alpha \beta \gamma = -\frac{d}{a} $$
Proof
Quadratic Equation
A quadratic equation whose leading coefficient is $a$ and whose roots are $\alpha$ and $\beta$ can be expressed as follows.
$$ a(x-\alpha) ( x-\beta)=0 $$
Expanding the left-hand side gives
$$ \begin{align*} a(x-\alpha) (x -\beta) =&\ a \left[ x^{2}-(\alpha +\beta)x+\alpha \beta \right] \\ =&\ ax^{2} -a(\alpha + \beta) x +a\alpha \beta \end{align*} $$
Therefore,
$$ b=-a(\alpha + \beta)\quad \& \quad a\alpha\beta=c $$
Arranging the above expression in terms of the roots gives
$$ \alpha+\beta=-\frac{b}{a}\quad \& \quad \alpha\beta= \frac{ c}{a} $$
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Cubic Equation
The proof method is similar to that of the quadratic equation. A cubic equation whose leading coefficient is $a$ and whose roots are $\alpha$, $\beta$, and $\gamma$ can be expressed as follows.
$$ a(x-\alpha) (x-\beta) (x-\gamma)=0 $$
Expanding the left-hand side gives
$$ \begin{align*} a(x-\alpha) (x-\beta) (x-\gamma) =&\a[x^{2}-(\alpha+\beta)x+\alpha \beta] (x-\gamma) \\ =&\ a\left[x^{2}-(\alpha + \beta + \gamma )x^{2}+(\alpha\beta + \beta\gamma + \gamma \alpha)x-\alpha\beta\gamma \right] \\ =&\ ax^{2}-a(\alpha + \beta + \gamma )x^{2}+a(\alpha\beta + \beta\gamma + \gamma \alpha)x-a\alpha\beta\gamma \end{align*} $$
Therefore,
$$ b=-a(\alpha+\beta + \gamma) \quad \& \quad c=a(\alpha\beta+\beta \gamma +\gamma \alpha) \quad \& \quad d=-a\alpha\beta\gamma $$
Arranging the above expression in terms of the roots gives
$$ \alpha + \beta + \gamma = -\frac{b}{a}\quad \& \quad \alpha\beta + \beta \gamma +\gamma \alpha = \frac{c}{a}\quad \& \quad \alpha \beta \gamma = -\frac{d}{a} $$
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Relationship Between Roots and Coefficients of Equations of Fourth Degree or Higher
From the above two formulas, it can be inferred that the relationship between the roots and coefficients of a quadratic equation $av^{4}+bx^{3}+cx^{2}+dx+e=0$ with roots $\alpha$, $\beta$, $\gamma$, and $\delta$ would be as follows, and it is indeed the case.
$$ \begin{align*} \alpha+\beta+\gamma+\delta= -\frac{b}{a}&& \& && \alpha\beta+\beta\gamma+\gamma\delta + \delta\alpha=\frac{c}{a} \\ \alpha\beta\gamma + \beta\gamma\delta+\gamma\delta\alpha+\delta\alpha\beta=-\frac{c}{a} && \& && \alpha\beta\gamma\delta=\frac{d}{a} \end{align*} $$
Of course, the same method applies to any polynomial of degree $n$. If the leading coefficient is $a,\ b,\ c,\ d,\ e,\ f,\ \cdots$, then
$$ \begin{align*} \text{모든 근의 합} =&-\frac{b}{a} \\ \text{모든 ‘두 근의 곱’의 합} =&\ \frac{c}{a} \\ \text{모든 ‘세 근의 곱’의 합} =&\ -\frac{d}{a} \\ \text{모든 ‘네 근의 곱’의 합} =&\ \frac{e}{a} \\ \text{모든 ‘다섯 근의 곱’의 합} =&\ -\frac{f}{a} \\ \vdots& \end{align*} $$