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Relationships between the Roots and Coefficients of Quadratic/Tertiary/nth Degree Equations 📂Abstract Algebra

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 ax2+bx+c=0ax^{2}+bx+c=0 to be α\alpha and β\beta. Then the following relationship holds true.

α+β=−ba&αβ=ca \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 ax3+bx2+cx+d=0ax^{3}+bx^{2}+cx+d=0 to be α\alpha, β\beta, and γ\gamma. Then the following relationship holds true.

α+β+γ=−ba&αβ+βγ+γα=ca&αβγ=−da \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 aa and whose roots are α\alpha and β\beta can be expressed as follows.

a(x−α)(x−β)=0 a(x-\alpha) ( x-\beta)=0

Expanding the left-hand side gives

a(x−α)(x−β)= a[x2−(α+β)x+αβ]= ax2−a(α+β)x+aαβ \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(α+β)&aαβ=c b=-a(\alpha + \beta)\quad \& \quad a\alpha\beta=c

Arranging the above expression in terms of the roots gives

α+β=−ba&αβ=ca \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 aa and whose roots are α\alpha, β\beta, and γ\gamma can be expressed as follows.

a(x−α)(x−β)(x−γ)=0 a(x-\alpha) (x-\beta) (x-\gamma)=0

Expanding the left-hand side gives

a(x−α)(x−β)(x−γ)=\a[x2−(α+β)x+αβ](x−γ)= a[x2−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ]= ax2−a(α+β+γ)x2+a(αβ+βγ+γα)x−aαβγ \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(α+β+γ)&c=a(αβ+βγ+γα)&d=−aαβγ 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

α+β+γ=−ba&αβ+βγ+γα=ca&αβγ=−da \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 av4+bx3+cx2+dx+e=0av^{4}+bx^{3}+cx^{2}+dx+e=0 with roots α\alpha, β\beta, γ\gamma, and δ\delta would be as follows, and it is indeed the case.

α+β+γ+δ=−ba&αβ+βγ+γδ+δα=caαβγ+βγδ+γδα+δαβ=−ca&αβγδ=da \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 nn. If the leading coefficient is a, b, c, d, e, f, ⋯a,\ b,\ c,\ d,\ e,\ f,\ \cdots, then

모든 근의 합=−ba모든 ‘두 근의 곱’의 합= ca모든 ‘세 근의 곱’의 합= −da모든 ‘네 근의 곱’의 합= ea모든 ‘다섯 근의 곱’의 합= −fa⋮ \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*}