# Vieta's Formula

**Vieta's formula** relates the coefficients of polynomials to the sums and products of their roots, as well as the products of the roots taken in groups.

For example, if there is a quadratic polynomial $f(x) = x^2+2x -15$, it will have roots of $x=-5$ and $x=3$, because $f(x) = x^2+2x-15=(x-3)(x+5)$. Vieta's formula can find the sum of the roots $\big( 3+(-5) = -2\big)$ and the product of the roots $\big(3 \cdot (-5)=-15\big)$ without finding each root directly. While this is fairly trivial in this specific example, Vieta's formula is extremely useful in more complicated algebraic polynomials with many roots or when the roots of a polynomial are not easy to derive. For some problems, Vieta's formula can serve as a shortcut to finding solutions quickly knowing the sums or products of their roots.

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## Vieta's Formula - Quadratic Equations

Let's start with a definition.

Vieta's Formula for Quadratics:Given $f(x) = ax^2+bx+c$, if the equation $f(x) = 0$ has roots $r_1$ and $r_2$, then

$r_1 + r_2 = -\frac{b}{a}, \quad r_1 r_2 = \frac{c}{a}.\ _\square$

The proof of this statement is given at the end of this section.

We immediately see how the coefficients of a quadratic help to determine the relationship of the roots.

If $\alpha$ and $\beta$ are the roots of the quadratic $x^2 - 4x + 9 = 0$, what are the values of

- $\alpha + \beta$
- $\alpha \beta$
- $\alpha^2 + \beta ^2 ?$

From Vieta's formula, we recognize that $\alpha + \beta = 4$.

From Vieta's formula, we recognize that $\alpha \beta = 9$.

Vieta's formula doesn't tell us the value of $\alpha^2 + \beta ^2$ directly. What we need to do, is to write $\alpha ^2 + \beta ^2$ in terms of $\alpha + \beta$ and/or $\alpha \beta$, and we can then substitute these values in. We have

$\begin{array} { l l } \alpha ^2 + \beta ^2 & = ( \alpha + \beta ) ^2 - 2 \alpha \beta \\ & = 4 ^ 2 - 2 \times 9 \\ & = - 2.\ _\square \\ \end{array}$

Note: In this question, the roots are the complex numbers $2 \pm \sqrt{5} i$. If we were to use these as the values of $\alpha$ and $\beta$ and then compute, we have a higher chance of making careless calculation mistakes. Vieta's formula offers us a simpler approach.

Can you quickly guess the roots of the quadratic $x^2-5x+6?$

If $p$ and $q$ are the roots of the given equation, then Vieta's formula tells us

$\begin{array}{c}&p+q=5, &pq=6,\end{array}$

and it's not hard to see that $2+3=5$ and $2\times 3=6.$ So the roots must be $2$ and $3$, and they indeed are. $_\square$

Given $\alpha$ and $\beta$ are the roots of the quadratic $a x^2 + bx + c = 0$, express $\dfrac{ b^2 - 4a c } { a^ 2 }$ in terms of $\alpha$ and $\beta$.

Vieta's formula gives us $\frac{ -b}{a} = \alpha + \beta$ and $\frac{c}{a} = \alpha \beta$. Substituting these in, we get

$\begin{array} { l l } \frac{ b^2 - 4ac } { a^2 } & = \frac{b^2} { a^2} - 4 \frac{c}{a} \\ & = ( \alpha + \beta) ^2 - 4 \alpha \beta \\ & = \alpha^2 + 2 \alpha \beta + \beta ^2 - 4 \alpha \beta \\ & = \alpha^2 - 2 \alpha \beta + \beta ^2 \\ & = ( \alpha - \beta ) ^ 2.\ _\square \end{array}$

You may recognize $b^2 - 4ac$ from the Quadratic Formula. In fact, we can show that

$\frac { - b \pm \sqrt{ b^2 - 4ac } } { 2a} = \frac{1}{2} \left( \frac{ -b}{a} \pm \sqrt{ \frac{ b^2 - 4ac } { a^2 } } \right) = \frac { ( \alpha + \beta ) \pm | \alpha - \beta | } { 2} = \alpha \text{ or } \beta,$

which proves the quadratic formula.

By the Remainder Factor Theorem, since the polynomial $f(x)$ has roots $r_1$ and $r_2$, it must have the form $f(x) = A ( x - r_1) ( x - r_2) = A x^2 - A ( r_1 + r_2) x + A r_1r_2$ for some constant $A$.

Comparing coefficients with $f(x) = ax^2 + bx + c$, we conclude that $a = A, b = -A ( r_1 + r_2)$ and $c = A r_1 r_2$. Hence, we get

$r_1 + r_2 = - \frac{ b}{a} , \quad r_1 r_2 = \frac{c}{a}.\ _\square$

## Vieta's Formula - Forming Quadratics

Let $p$ and $q$ be the real roots of the monic quadratic equation

$\begin{array}{c}&x^2+bx+c=0, &(b,c)\in\mathbb{R}^2.\end{array}$

Why monic? Because we can always divide the whole equation by its leading coefficient to get a monic version of it. So what can we say about $b,c$? Since $p$ and $q$ are the roots of this equation, we can factorize the equation as

$x^2+bx+c\equiv \left(x-p\right)\left(x-q\right).$

Expanding the right side and rearranging, we find

$x^2+bx+c\equiv x^2-(p+q)x+pq.$

Since two polynomials are equal if and only if their coefficients are equal, by equating the coefficients we get

$\begin{array}{c}&b=-(p+q), &c=pq.\end{array}$

This is the so-called Vieta's formula for a quadratic polynomial. It can be similarly extended to polynomials of higher degree.

The roots can be generalized to include complex numbers. That is, given two complex numbers $p$ and $q$, we can always construct a monic quadratic whose roots are $p$ and $q$. More specifically, the quadratic will be

$x^2-(p+q)x+pq=0.$

When will both the coefficients be real? To find the answer, set $p=p_1+p_2 i$ and $q=q_1+q_2 i$. Then the coefficients are

$\begin{array}{c}&b=-\left(p_1+q_1\right)-\left(p_2+q_2\right)i, &c=\left(p_1q_1-p_2q_2\right)+\left(p_1q_2+p_2q_1\right)i.\end{array}$

For $b$ to be real, we need $p_2+q_2=0 \Rightarrow p_2=-q_2$. For $c$ to be real, we need

$p_1q_2+p_2q_1=0 \Rightarrow q_2\left(p_1-q_1\right)=0,$

implying either $q_2=0=p_2$ or $p_1=q_1$. Hence either we need both $p$ and $q$ real, or $p$ and $q$ are complex conjugates of each other.

Find a quadratic whose roots are $2$ and $5$.

Let the quadratic be $x^2+bx+c$, where we wish to find $b$ and $c$. Then Vieta's formula tells us that

$\begin{array}{c}&b=-(2+5)=-7, &c=2\times 5 = 10.\end{array}$

Therefore the desired quadratic is $x^2-7x+10.$ $_\square$

Find a quadratic whose roots are $3+2i$ and $3-2i$.

Let the desired quadratic be $x^2+bx+c$, where we wish to find $b$ and $c$. Then Vieta's formula tells us

$\begin{array}{c}&b=-\left[\left(3+2i\right)+\left(3-2i\right)\right]=-6, &c=\left(3+2i\right)\left(3-2i\right)=13.\end{array}$

So the desired quadratic is $x^2-6x+13.$ $_\square$

Note: Since the roots were complex conjugates of each other, the coefficients of the quadratic became real.

Solve the system of equations

$\begin{array}{c}&a+b = 7, &ab = 10.\end{array}$

By Vieta's Formula, we know that $a$ and $b$ are the roots of the equation $x^2 - 7x + 10 = 0$. Since we can factorize it as $(x-2) ( x-5) =0$, we get that $\{ a,b \} = \{2, 5 \}.$ $_\square$

## Generalization to Higher Degree Polynomials

Consider a quadratic equation with complex coefficients and roots $r_1$, $r_2:$

$a_2x^2+a_1x+a_0 = a_2(x-r_1)(x-r_2).$

By comparing coefficients, we can see that

$\begin{array}{c}&r_1+r_2=-\frac{a_1}{a_2}, &r_1r_2=\frac{a_0}{a_2}.\end{array}$

This gives a relationship between the roots of a polynomial and the coefficients of the polynomial. Generalizing this idea for a polynomial of degree $n$, we have the following formula:

Vieta's Formula:Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ be a polynomial with complex coefficients and degree $n$, having complex roots $r_n, r_{n-1}, \ldots, r_1$. Then for any integer $0 \leq k \leq n,$

$\sum\limits_{1\leq i_1 < i_2 < \cdots < i_k \leq n} r_{i_1}r_{i_2}\cdots r_{i_k} = (-1)^k \frac{a_{n-k}}{a_n}.\ _\square$

The expressions on the left-hand side of Vieta's formula are the elementary symmetric functions of $r_1, r_2, \ldots r_n$. The proof of Vieta's formula follows by comparing coefficients in the equation

$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = a_n (x-r_1)(x-r_2)(x-r_3) \cdots (x-r_n).$

As in the quadratic case, Vieta's formula gives an equation to find the sum of roots:

$\sum_{i=1}^n r_i = - \frac{a_{n-1}}{a_n}.$

Similarly, we have the following equation for the product of roots:

$r_1 r_2 \cdots r_n = (-1)^n \frac{a_{0}}{a_n}.$

Vieta's formula gives relationships between polynomial roots and coefficients that are often useful in problem-solving.

Suppose $k$ is a number such that the cubic polynomial $P(x) = -2x^3 + 48 x^2 + k$ has three integer roots that are all prime numbers. How many possible distinct values are there for $k?$

Let $p, q,$ and $r$ denote the three integer roots of $P(x)$. Then by Vieta's formula, we have $pq + qr + pr = 0$, but since $p, q$ and $r$ are prime, each of them are strictly greater than 1, and hence no such $P(x)$ exists. $_\square$

[1968 Putnam Exam] Find all polynomials $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ such that $a_i=\pm 1$ for all $0 \leq i \leq n-1$ satisfying the condition that all roots of $P(x)$ are real.

If $r_1, r_2, \cdots, r_n$ are the real roots of $P(x)$, then Vieta's formula implies

$\sum_{i=1}^n r_i = -a_{n-1} \text{ and } \sum_{1 \leq i<j \leq n} r_i r_j = a_{n-2},$

which implies

$\begin{aligned} \sum_{i=1}^n r_i^2 &= \left( \sum_{i=1}^n r_i \right)^2 - 2 \left(\sum_{1 \leq i<j \leq n} r_i r_j\right)\\ & = a_{n-1}^2 -2 a_{n-2}\\ & \leq 3 .\\ \end{aligned}$

Note that this also shows that $a_{n-1}^2 -2 a_{n-2} \geq 0$ since the left hand side is the sum of squares of real numbers. Since $a_{n-1} = \pm 1$, this implies $a_{n-2} = -1$. Also by Vieta's formula, $\left( r_1 r_2 \cdots r_n \right)^2 = 1$. Then by the AM-GM inequality, we have $\sum_{i=1}^n r_i^2 \geq n$, implying $n \leq 3$. We can then enumerate all such polynomials to find the solutions are $x \pm 1$, $x^2 \pm x -1$, and $x^3 - x \pm (x^2-1)$. $_\square$

Let $r_1, r_2$ and $r_3$ be the roots of the polynomial $5x^3 -11x^2+7x+3$. Evaluate $r_1^3+r_2^3+r_3^3.$

In this problem, the application of Vieta's formulas is not immediately obvious, and the expression has to be transformed. From factorization 4, we have

$r_1^3+r_2^3+r_3^3 = 3r_1r_2r_3 + (r_1+r_2+r_3)\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \right].$

Now we only need to know how to calculate $r_1^2+r_2^2+r_3^2$. Again, from factorization, we have

$r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1),$

which allows us to conclude that

$\begin{array} { l l } r_1^3+r_2^3+r_3^3 & = 3r_1r_2r_3 + (r_1+r_2+r_3) \left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \right] \\ & =3r_1r_2r_3 + (r_1+r_2+r_3 ) \left[ (r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1) \right] \\ & =3\times \left(-\frac{3}{5}\right) +\left(\frac{11}{5}\right) \left[ \left(\frac{11}{5} \right)^2-3\times \frac{7}{5} \right] \\ & = -\frac{49}{125}. \ _\square \end{array}$

## Vieta's Formula Problem Solving - Basic

Find all triples of complex numbers which satisfy the following system of equations:

$\begin{aligned} a + b + c & = 0\\ ab + bc + ca & = 0 \\ abc & = 0. \end{aligned}$

Any $3$ numbers $(a,b,c)$ can be considered as the roots of the monic cubic polynomial

$P(x) = (x-a)(x-b)(x-c).$

Applying Vieta's formula to the polynomial, we get $P(x)=x^3$. Hence, the only root of $P(x)$ is $x=0$ (with multiplicity $3$), which implies that $(0,0,0)$ is the only triple of complex numbers that satisfies the system of equations. $_\square$

Let $r_1, r_2$ and $r_3$ be the roots of the polynomial $5x^3 -11x^2+7x+3$. Evaluate

$r_1(1+r_2+r_3) + r_2(1+r_3+r_1) + r_3(1+r_1+r_2) .$

The expression is equal to $r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1)$. By Vieta's Formula, we know that

$\begin{array}{c}&r_1+r_2+r_3=-\frac{-11}{5}=\frac{11}{5}, &r_1r_2+r_2r_3+r_3r_1=\frac{7}{5}, &r_1r_2r_3 = - \frac{3}{5}, \end{array}$

so

$\begin{aligned} r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1) &= \frac{11}{5}+2\times \frac{7}{5} \\ &= \frac{25}{5}\\ &=5.\ _\square \end{aligned}$

Note: A common approach would be to try and find each root of the polynomial, especially since we know that one of the roots must be real (why?). However, this is not necessarily a viable option, since it is hard for us to determine what the roots actually are.

## Vieta's Formula Problem Solving - Intermediate

Let $r_1, r_2$ and $r_3$ be the roots of the polynomial $5x^3 -11x^2+7x+3$. Evaluate $r_1^3+r_2^3+r_3^3.$

In this problem, the application of Vieta's formulas is not immediately obvious, and the expression has to be transformed. From factorization, we have

$r_1^3+r_2^3+r_3^3 = 3r_1r_2r_3 + (r_1+r_2+r_3)\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \right].$

Now we only need to know how to calculate $r_1^2+r_2^2+r_3^2$. Again, from factorization, we have

$r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1),$

which allows us to conclude that

$\begin{array} { l l } r_1^3+r_2^3+r_3^3 & = 3r_1r_2r_3 + (r_1+r_2+r_3) \left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \right] \\ & =3r_1r_2r_3 + (r_1+r_2+r_3 ) \left[ (r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1) \right] \\ & =3\times \left(-\frac{3}{5}\right) +\left(\frac{11}{5}\right) \left[ \left(\frac{11}{5} \right)^2-3\times \frac{7}{5} \right] \\ & = -\frac{49}{125}.\ _\square \end{array}$

If $\alpha, \beta$ are the roots to the equation $x^2 +2x+3 = 0$, what is the quadratic equation whose roots are $\left(\alpha - \dfrac{1}{\alpha} \right)^2$ and $\left( \beta - \dfrac{1}{\beta} \right) ^ 2?$

[ARML 2012, Team Problems, #6]

The zeros of $f(x) = x^6 + 2x^5 + 3x^4 + 5x^3 + 8x^2 + 13x + 21$ are distinct complex numbers. Compute the average value of $A + BC + DEF$ over all possible permutations $(A, B, C, D, E, F)$ of these six numbers.

## Vieta's Formula Problem Solving - Advanced

[IMO shortlist]

Determine all real values of the parameter $a$ for which the equation$16x^4-ax^3+(2a+17)x^2-ax+16 = 0$

has exactly four distinct real roots that form a geometric progression.

Suppose that $a$ satisfies the requirements of the problem and that $x$, $qx$, $q^2x$, $q^3x$ are the roots of the given equation. Then $x \neq 0$ and we may assume that $|q|>1$, so that $|x|<|qx|<|q^2x|<|q^3x|$. Notice that the coefficients are symmetric, namely the first coefficient is the same as the fifth one, the second is the same as the fourth and the third is the same as the third. It guarantees us that if $\alpha$ is a root, then its reciprocal (which is $\alpha ^{-1} = \frac{1}{\alpha}$) will also be a root. Hence, $\frac{1}{x} = q^3 x,$ so $q = x^{-\frac{2}{3} }$ and the roots are $x, x^{\frac{1}{3}}, x^{\frac{-1}{3}}, x^{-1}$.

Now, by Vieta's formula we have $x+x^{\frac{1}{3}}+x^{\frac{-1}{3}}+x^{-1}=\frac{a}{16}$ and $x^{\frac{4}{3}}+x^{\frac{2}{3}}+1 + 1+x^{\frac{-2}{3}}+x^{\frac{-4}{3}}=\frac{2a+17}{16}$. On setting $z =x^{\frac{1}{3}}+x^{\frac{-1}{3}}$ these equations become

$\begin{array}{c}&z^3-2z=\frac{a}{16}, &(z^2-2)^2+z^2-2=\frac{2a+17}{16}.\end{array}$

Substituting $a=16(z^3-2z)$ in the second equation leads to $z^4-2z^3-3z^2+4z+\frac{15}{16} = 0$. We observe that this polynomial factors as $\left(z+\frac{3}{2}\right)\left(z-\frac{5}{2}\right)\left(z^2-z-\frac{1}{4}\right)$. Since $|z|=\left|x^{\frac{1}{3}}+x^{-\frac{1}{3}}\right|\geq 2$, the only viable value is $z=\frac{5}{2}$. Consequently $a=170.$

Rearranging the equation, we get $16 \left(x^2 + \frac{1}{x^2}\right) - 170\left(x + \frac{1}{x}\right) + 357 = 0$. To simplify it, we can call $y = x + \frac{1}{x}$ and therefore $y^2 = x^2 + 2 + \frac{1}{x^2}$, thus getting a new form

$16 (y^2 - 2) - 170y + 357 = 0 \implies 16y^2 - 170y +325 = 0,$

whose roots are $\frac{5}{2}$ and $\frac{65}{8}$. We have to plug the two back into $y = x + \frac{1}{x}$, leading us to two more quadratics, getting finally $\begin{array}{c}&\dfrac{1}{8}, &\dfrac{1}{2}, &2, &8.\end{array}\ _\square$

(Created by Mark Hennings)

Show that$\sum_{j=1}^n \cot^2\Big(\tfrac{j \pi}{2n+1}\Big) \; = \; \tfrac13n(2n-1)$

for any integer $n \ge 1$.

(Solution by Aditya Sharma)

Let us start with De-Moivre's theorem for any positive integer $m:$

$\begin{aligned} (\cos x+i\sin x)^m&=\cos mx+i\sin mx\\ \frac{\cos mx+i\sin mx}{\sin^m x}&=(\cot x+i)^m\\ &=\cot^m x+\binom{m}{1}\cot^{m-1}(x)i+\cdots+\binom{m}{m-1}\cot(x)i^{m-1} + i^m. \end{aligned}$

We can group the RHS as follows since we have $i^2=-1:$

$\text{RHS} =\left(\cot^m x-\binom{m}{2}\cot^{m-2} x+\cdots\right)+i\left(\binom{m}{1}\cot^{m-1}x-\binom{m}{3}\cot^{m-3} x+\cdots\right).$

Equating the imaginary parts on the LHS and RHS, we get

$\binom{m}{1}\cot^{m-1}x-\binom{m}{3}\cot^{m-3}x+\cdots = \frac{\sin mx}{\sin^m x}.$

Now if we let $m=2n+1$ and $x=\frac{j\pi}{m}$ for any $j=1,2,\ldots, m$, then the equation reduces to

$\binom{2n+1}{1}\cot^{2n}x-\binom{2n+1}{3}\cot^{2n-2} x+\cdots = 0$

since $\sin mx=\sin j\pi=0.$

Let $\cot^2 x = u,$ then the equation is in $u$ and its sum of roots is given by $\displaystyle \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}}$ as we know from Vieta's formula. Since $x=\frac{j\pi}{2n+1},$

$\begin{aligned} \sum_{j=1}^{n}\cot^2 \frac{j\pi}{2n+1} &= \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}}\\ &= \frac{1}{6}2n(2n-1)\\ &= \frac{1}{3}n(2n-1). \ _\square \end{aligned}$

What is

$\sum_{k=1}^n \cot^4\Big(\tfrac{k \pi}{2n+1}\Big)?$

Let $w$ be any primitive $(2n+1)^\text{th}$ root of unity. Then, $P(x)=\dfrac{x^{2n+1}-1}{x-1}$ will have roots $w,w^2,\ldots,w^{2n}$. We also know that $\sin\left(\frac{\pi k}{2n+1}\right)=\frac{w^{k/2}-w^{-k/2}}{2i}$ and $\cos\left(\frac{\pi k}{2n+1}\right)=\frac{w^{k/2}+w^{-k/2}}{2}$. Then $\cot\left(\frac{\pi k}{2n+1}\right)=\frac{i(w^k+1)}{w^k-1}$. So the sum $S$ (say) is

$S=\sum_{k=1}^{n}\left(\dfrac{w^k+1}{w^k-1}\right)^4.$

Now, we will do a polynomial transformation of roots by doing $y=\frac{x+1}{x-1}$, then $x=\frac{y+1}{y-1}$. Substitute that in $P(x)$:

$P(x)=\dfrac{\left(\dfrac{y+1}{y-1}\right)^{2n+1}-1}{\dfrac{y+1}{y-1}-1}.$

Simplifying and making it a polynomial, we get

$P(y)=(y+1)^{2n+1}-(y-1)^{2n+1}.$

Expand and simplify again:

$P(y)=2\binom{2n+1}{1}\big(y^2\big)^n+2\binom{2n+1}{3}\big(y^2\big)^{n-1}+2\binom{2n+1}{5}\big(y^2\big)^{n-2}+\cdots.$

Finally, by Vieta's formulas, our sum is

$\begin{aligned} S=\sum_{k=1}^n y_k^4 &=\left(\sum_{k=1}^{n}y_k^2\right)^2-2\left(\sum_{1\leq j<k\leq n}y_j^2 y_k^2\right)\\\\ &=\left(\dfrac{-2\binom{2n+1}{3}}{2\binom{2n+1}{1}}\right)^2-2\left(\dfrac{2\binom{2n+1}{5}}{2\binom{2n+1}{1}}\right)\\\\ &=\dfrac{n^2(2n-1)^2}{9}-\dfrac{n(2n-1)(n-1)(2n-3)}{15}\\\\ &=\dfrac{n(2n-1)(4n^2+10n-9)}{45}. \ _\square \end{aligned}$

## Vieta Root Jumping

Main Article: Vieta Root Jumping

Vieta jumping is a nickname for a particular kind of descent method that has become quite popular in higher level math olympiad number theory problems. Like other instances of descent, it occurs when you have to solve a Diophantine equation (or system of equations, congruences or inequalities) whose solutions have some recursive structure.