Viète's theorem and its applications

François Viète (1540–1603) was a lawyer, a royal counsellor to King Henry III of France and later worked for King Henry IV. He decoded secret letters sent to King Philip II of Spain, so that Henry IV learnt of his enemy's military plans. In mathematics, he introduced the first systematic algebraic notation: vowels for the unknowns and consonants for the known numbers, which let him write general rules for every equation (MacTutor, University of St Andrews). The link between roots and coefficients that we study today bears his name.

Minh tells the class: "The equation $x^2-7x+12=0$ has roots $3$ and $4$. I knew at once, without working out $\Delta$, because $3+4=7$ and $3\cdot4=12$." Was Minh lucky, or is there a rule behind it? And if there is, what can that rule do for us?

Viète's theorem

Explore: the sum and product of the roots

We solve a few equations (with the quadratic formula) and work out the sum and product of the roots:

In the second row the sum is $\dfrac72=\dfrac{-(-7)}{2}$ and the product is $\dfrac32$: the sum is always $-\dfrac{b}{a}$ and the product is always $\dfrac{c}{a}$.

Why? When $\Delta\geq0$, add and multiply the two roots $\dfrac{-b\pm\sqrt\Delta}{2a}$ (lesson quadratic equations); the square roots cancel:

$$x_1+x_2=\frac{-2b}{2a}=-\frac ba,\qquad x_1x_2=\frac{(-b)^2-(\sqrt\Delta)^2}{4a^2}=\frac{b^2-(b^2-4ac)}{4a^2}=\frac ca.$$

Going deeper. If $x_1$, $x_2$ are the roots, then $(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$. Comparing this with $x^2+\dfrac bax+\dfrac ca$ (the equation divided by $a$), we see that Viète's theorem is very natural. But the theorem only talks about roots that exist: always check that $\Delta\geq0$ (or that $a$ and $c$ have opposite signs) before using it.

From the sum $S$ and the product $P$ we can work out many "symmetric" expressions (swapping $x_1$ and $x_2$ does not change their value), for example $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=S^2-2P$.

Finding the roots mentally

Substituting $x=1$ into $ax^2+bx+c$ gives $a+b+c$; substituting $x=-1$ gives $a-b+c$. Together with the product $x_1x_2=\dfrac ca$, we find the other root at once.

Part c) shows the power of this method: with the quadratic formula you would have to work out $\Delta=2027^2-4\cdot2026=2025^2$.

Finding two numbers from their sum and product

Viète's theorem also works backwards: if two numbers $u$, $v$ have $u+v=S$ and $uv=P$, then $(x-u)(x-v)=x^2-Sx+P$, so $u$ and $v$ are the roots of $x^2-Sx+P=0$.

The signs of the roots

Without solving, $S$ and $P$ tell us the signs of the two roots (when the equation has roots):

• $P<0$: the roots have opposite signs (then $a$ and $c$ have opposite signs, so the equation certainly has two roots);
• $\Delta\geq0$, $P>0$, $S>0$: both roots are positive;
• $\Delta\geq0$, $P>0$, $S<0$: both roots are negative.

The reason: a negative product means the numbers have different signs; a positive product means the same sign, and the sum shows which sign.

Connections. Viète's theorem follows directly from the quadratic formula and gives a new point of view: instead of finding each root, we work with the pair $(S, P)$. This point of view is used a lot in solving word problems by forming equations (rectangle problems), in factorising $ax^2+bx+c=a(x-x_1)(x-x_2)$, and it is a standard question type in the grade 10 entrance exam.

Key terms

Summary

• If $\Delta\geq0$: $x_1+x_2=-\dfrac ba$, $x_1x_2=\dfrac ca$.
• $a+b+c=0$: roots $1$ and $\dfrac ca$; $a-b+c=0$: roots $-1$ and $-\dfrac ca$.
• Two numbers with sum $S$ and product $P$ are the roots of $x^2-Sx+P=0$ (need $S^2-4P\geq0$).
• $x_1^2+x_2^2=S^2-2P$; $\dfrac1{x_1}+\dfrac1{x_2}=\dfrac SP$.
• $P<0$: opposite signs; $P>0$, $S>0$: both positive; $P>0$, $S<0$: both negative (when $\Delta\geq0$).

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Mini test