Further Maths/Lessons/Roots of Polynomials/Roots of a Quadratic Equation
Core Pure · Roots of Polynomials · Year 1

Roots of a Quadratic Equation

The sum and product of a quadratic's roots are hiding in plain sight in its coefficients — no need to solve the equation to know a great deal about its roots.

16 min read 5 worked examples

Sum and product of roots

Theorem 10.1 — Vieta's formulae for a quadratic

If $\alpha,\beta$ are the roots of $ax^2+bx+c=0$ ($a\neq0$), then $$\alpha+\beta=-\frac{b}{a}, \qquad \alpha\beta=\frac{c}{a}$$

ProofVieta's formulae by comparing factor form
1
Since $\alpha,\beta$ are roots, $ax^2+bx+c\equiv a(x-\alpha)(x-\beta)$.
A quadratic with leading coefficient $a$ and roots $\alpha,\beta$ factorises this way.
2
Expand the right side: $a(x-\alpha)(x-\beta)=a\big[x^2-(\alpha+\beta)x+\alpha\beta\big]=ax^2-a(\alpha+\beta)x+a\alpha\beta$.
Standard expansion of two linear factors.
3
Compare coefficients of $x$: $b=-a(\alpha+\beta) \Rightarrow \alpha+\beta=-\dfrac{b}{a}$.
Two polynomials are identically equal only if corresponding coefficients match.
4
Compare constant terms: $c=a\alpha\beta \Rightarrow \alpha\beta=\dfrac{c}{a}$.
Same principle applied to the constant term.
∎
Example 1 · Finding sum and product directly
Find the sum and product of the roots of $3x^2-7x+2=0$.

$\alpha+\beta=-\dfrac{-7}{3}=\dfrac73$, $\quad \alpha\beta=\dfrac23$.

Symmetric functions of the roots

Method 10.2 — evaluating symmetric expressions

Many expressions in $\alpha,\beta$ can be rewritten purely in terms of $\alpha+\beta$ and $\alpha\beta$ using standard identities: $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta,$$ $$\frac{1}{\alpha}+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta},$$ $$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$$ This means such expressions can be evaluated without ever finding $\alpha$ and $\beta$ individually.

Example 2 · Evaluating α² + β²
The roots of $2x^2-5x+3=0$ are $\alpha$ and $\beta$. Find $\alpha^2+\beta^2$.

$\alpha+\beta=\dfrac52$, $\alpha\beta=\dfrac32$.

$\alpha^2+\beta^2=\left(\dfrac52\right)^2-2\left(\dfrac32\right)=\dfrac{25}{4}-3=\dfrac{25}{4}-\dfrac{12}{4}=\dfrac{13}{4}$.

Example 3 · Forming a new equation from squared roots
The roots of $x^2-6x+4=0$ are $\alpha$ and $\beta$. Find the quadratic equation with roots $\alpha^2$ and $\beta^2$.

$\alpha+\beta=6$, $\alpha\beta=4$.

New sum: $\alpha^2+\beta^2=6^2-2(4)=36-8=28$.

New product: $\alpha^2\beta^2=(\alpha\beta)^2=16$.

New equation: $x^2-28x+16=0$.

Example 4 · Using given information to find an unknown coefficient
The roots $\alpha,\beta$ of $x^2+px+8=0$ satisfy $\alpha^2+\beta^2=17$. Find the possible values of $p$.

$\alpha+\beta=-p$, $\alpha\beta=8$.

$\alpha^2+\beta^2=(-p)^2-2(8)=p^2-16=17 \Rightarrow p^2=33 \Rightarrow p=\pm\sqrt{33}$.

Common mistake

$\alpha^2+\beta^2 \neq (\alpha+\beta)^2$ — students frequently forget the $-2\alpha\beta$ correction term. Always derive this identity from $(\alpha+\beta)^2=\alpha^2+2\alpha\beta+\beta^2$ if you're unsure, rather than trying to recall it directly.

Method 10.3 — the sum-of-cubes identity

A similar trick handles $\alpha^3+\beta^3$: expand $(\alpha+\beta)^3$ and isolate the cubic terms. $$(\alpha+\beta)^3=\alpha^3+3\alpha^2\beta+3\alpha\beta^2+\beta^3=\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$$ Rearranging gives $$\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$$ As before, this is entirely in terms of the sum and product, so no need to solve the quadratic.

Example 5 · Sum of cubes and reversing the process
The roots of $x^2-3x+5=0$ are $\alpha$ and $\beta$. (a) Find $\alpha^3+\beta^3$. (b) A different quadratic $ax^2+bx+c=0$ has roots $\alpha=4$ and $\beta=-\tfrac13$. Find integer values for $a$, $b$ and $c$.

(a) $\alpha+\beta=3$, $\alpha\beta=5$, so $\alpha^3+\beta^3=3^3-3(5)(3)=27-45=-18$.

(b) $\alpha+\beta=4-\dfrac13=\dfrac{11}{3}$ and $\alpha\beta=4\times\left(-\dfrac13\right)=-\dfrac43$.

Using $\alpha+\beta=-\dfrac ba$ and $\alpha\beta=\dfrac ca$, choose $a=3$: then $-\dfrac ba=\dfrac{11}{3}\Rightarrow b=-11$, and $\dfrac ca=-\dfrac43\Rightarrow c=-4$.

So $3x^2-11x-4=0$ has $a=3$, $b=-11$, $c=-4$ — any non-zero multiple of this equation also works.

Exam-style questions

Question 1 6 marks

The roots of $3x^2-7x+2=0$ are $\alpha$ and $\beta$. Find the value of (a) $\alpha^3+\beta^3$ using $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$, (b) $\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}$.

$\alpha+\beta=\dfrac73$, $\alpha\beta=\dfrac23$ (B1)

(a) $\left(\dfrac73\right)^3-3\left(\dfrac23\right)\left(\dfrac73\right)=\dfrac{343}{27}-\dfrac{14}{3}=\dfrac{343}{27}-\dfrac{126}{27}=\dfrac{217}{27}$ (M1, A1, A1)

(b) $\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}=\dfrac{\alpha^2+\beta^2}{(\alpha\beta)^2}=\dfrac{(7/3)^2-2(2/3)}{(2/3)^2}=\dfrac{49/9-4/3}{4/9}=\dfrac{37/9}{4/9}=\dfrac{37}{4}$ (M1, A1)

Question 2 6 marks

The roots of $2x^2+kx+18=0$ are $\alpha$ and $\beta$, where $\alpha=3\beta$. Find the possible values of $k$.

$\alpha\beta=9 \Rightarrow 3\beta\cdot\beta=9 \Rightarrow \beta^2=3 \Rightarrow \beta=\pm\sqrt3$ (M1, A1)

$\alpha+\beta=4\beta=-\dfrac{k}{2} \Rightarrow k=-8\beta$ (M1, A1)

$k=-8\sqrt3$ or $k=8\sqrt3$ (A1, A1)

Question 3 7 marks

Given that $\alpha$ and $\beta$ are the roots of $x^2-4x+7=0$, find a quadratic equation with integer coefficients whose roots are $\alpha+\dfrac{1}{\beta}$ and $\beta+\dfrac1\alpha$.

$\alpha+\beta=4$, $\alpha\beta=7$. (B1)

New sum: $\left(\alpha+\dfrac1\beta\right)+\left(\beta+\dfrac1\alpha\right)=(\alpha+\beta)+\dfrac{\alpha+\beta}{\alpha\beta}=4+\dfrac47=\dfrac{32}{7}$ (M1, A1)

New product: $\left(\alpha+\dfrac1\beta\right)\left(\beta+\dfrac1\alpha\right)=\alpha\beta+1+1+\dfrac{1}{\alpha\beta}=7+2+\dfrac17=\dfrac{64}{7}$ (M1, A1)

Equation: $x^2-\dfrac{32}{7}x+\dfrac{64}{7}=0$, or scaling by 7: $7x^2-32x+64=0$ (M1, A1)

← Previous
Sums of Squares & Cubes