Roots of Quadratic, Cubic & Quartic Equations
The coefficients of a polynomial secretly encode the sum and products of its roots — without ever needing to solve the equation itself.
Roots of a quadratic equation
If $\alpha,\beta$ are the roots of $ax^2+bx+c=0$ ($a\neq0$), then $$\alpha+\beta=-\frac{b}{a}, \qquad \alpha\beta=\frac{c}{a}.$$
Since $\alpha,\beta$ are roots, the quadratic factorises as $a(x-\alpha)(x-\beta)$. Expanding: $a(x-\alpha)(x-\beta)=a\big[x^2-(\alpha+\beta)x+\alpha\beta\big]=ax^2-a(\alpha+\beta)x+a\alpha\beta$. Comparing this with $ax^2+bx+c$ term by term gives $-a(\alpha+\beta)=b$ and $a\alpha\beta=c$, which rearrange to the theorem.
Roots of a cubic equation
If $\alpha,\beta,\gamma$ are the roots of $ax^3+bx^2+cx+d=0$, then $$\alpha+\beta+\gamma=-\frac{b}{a},$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a},$$ $$\alpha\beta\gamma=-\frac{d}{a}.$$
The same idea applies: expand $a(x-\alpha)(x-\beta)(x-\gamma)$ and compare coefficients with $ax^3+bx^2+cx+d$. The three elementary symmetric combinations of the roots — their sum, the sum of their pairwise products, and their product — appear as the coefficients in turn, alternating in sign.
Roots of a quartic equation
If $\alpha,\beta,\gamma,\delta$ are the roots of $ax^4+bx^3+cx^2+dx+e=0$, then $$\sum\alpha=-\frac{b}{a},\quad \sum\alpha\beta=\frac{c}{a},\quad \sum\alpha\beta\gamma=-\frac{d}{a},\quad \alpha\beta\gamma\delta=\frac{e}{a}$$ where $\sum\alpha\beta$ means the sum of all $\binom{4}{2}=6$ pairwise products, and $\sum\alpha\beta\gamma$ the sum of all $\binom{4}{3}=4$ triple products.
The pattern continues exactly as before: the sign alternates with each new elementary symmetric function, and each is divided by the leading coefficient $a$.
$\alpha+\beta=\dfrac52$, $\alpha\beta=\dfrac32$. Using $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$:
$\alpha+\beta+\gamma=6$, $\alpha\beta+\beta\gamma+\gamma\alpha=11$. Using $(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)$:
Product of roots: $\alpha\beta\gamma=-\dfrac{24}{2}=-12$, so $2\times3\times k=-12 \Rightarrow k=-2$.
Sum of roots: $\alpha+\beta+\gamma=-\dfrac{p}{2}$, so $2+3-2=-\dfrac{p}{2} \Rightarrow 3=-\dfrac{p}{2}\Rightarrow p=-6$.
Sum of pairwise products: $\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{q}{2}$, so $(2)(3)+(3)(-2)+(-2)(2)=6-6-4=-4=\dfrac{q}{2}\Rightarrow q=-8$.
The sign pattern alternates: $+,-,+,-,\ldots$ for sum, pairwise-sum, triple-sum, and so on — but whether the coefficient ratio $\left(\dfrac{b}{a}, \dfrac{c}{a}, \ldots\right)$ itself carries a minus sign depends on this alternation. Always re-derive the sign for the specific relation you need (as in Theorem 2.1's proof) rather than trying to memorise it blindly — a mixed-up sign is the most common way marks are lost on this topic.
Exam-style questions
Question 1 6 marks
The roots of $3x^2-7x+2=0$ are $\alpha,\beta$. Given that $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$, find the exact value of $\alpha^3+\beta^3$.
$\alpha+\beta=\dfrac73$, $\alpha\beta=\dfrac23$ (B1, B1)
$\alpha^3+\beta^3 = \left(\dfrac73\right)^3-3\left(\dfrac23\right)\left(\dfrac73\right) = \dfrac{343}{27}-\dfrac{14}{3} = \dfrac{343}{27}-\dfrac{126}{27}=\dfrac{217}{27}$ (M1, M1, A1, A1)
Question 2 6 marks
The cubic equation $2x^3+px^2+qx+24=0$ has roots $2$, $3$ and $k$. Find the values of $k$, $p$ and $q$.
Product: $2\times3\times k = -\dfrac{24}{2}=-12 \Rightarrow k=-2$ (M1, A1)
Sum: $2+3+(-2)=-\dfrac{p}{2}\Rightarrow 3=-\dfrac{p}{2}\Rightarrow p=-6$ (M1, A1)
Pairwise sum: $(2)(3)+(3)(-2)+(-2)(2)=-4=\dfrac{q}{2}\Rightarrow q=-8$ (M1, A1)