# Vieta's Formula

Consider a quadratic equation with complex coefficients and roots $r_1$, $r_2$ : $a_2(x-r_1)(x-r_2) = a_2x^2+a_1x+a_0.$ By comparing coefficients, we can see that $r_1+r_2=-\frac{a_1}{a_2}, r_1r_2=\frac{a_0}{a_2}.$

Generalizing this idea for a polynomial of degree $n$, we have

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 $r_n, r_{n-1}, \cdots, r_1$. Then for any integer $0 \leq k \leq n$:

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

This follows immediately by comparing coefficients on the polynomial $\prod ( x- r_i )$. We leave the proof to the reader.

## Worked Examples

### 1. Find all triples of complex numbers which satisfy the system of equations:

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

Solution: 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.

### 2. Let $r_1, r_2$ and $r_3$ be the roots of the polynomial $5x^3 -11x^2+7x+3$. Evaluate $r_1(1+r_2+r_3) + r_2(1+r_3+r_1) + r_3(1+r_1+r_2) .$

Solution: 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

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

so $r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1) = \frac{11}{5}+2\times \frac{7}{5} = \frac{25}{5}=5$.

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.

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

Solution: 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 that

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

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

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

so this allows us to conclude that

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

### 4. [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.

Solution: 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

$z^3-2z=\frac{a}{16}, (z^2-2)^2+z^2-2=\frac{2a+17}{16}.$

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

Rearranging the equation, we get $16 (x^2 + \frac{1}{x^2}) - 170(x + \frac{1}{x}) + 357 = 0$. 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 $16 (y^2 - 2) - 170y + 357 = 0 \Rightarrow 16y^2 - 170y +325 = 0$, 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 $\boxed{\dfrac{1}{8}, \dfrac{1}{2}, 2, 8}$.

## Test Yourself

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

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.

3. (Created by Mark Hennings. This is HARD) Show that $\sum_{j=1}^n \cot^2\Big(\tfrac{j \pi}{2n+1}\Big) \; = \; \tfrac13n(2n-1)$ for any integer $n \ge 1$. What is $\sum_{j=1}^n \cot^4\Big(\tfrac{j \pi}{2n+1}\Big) \qquad \mbox{?}$

Note by Calvin Lin
6 years, 6 months ago

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The solution to example two states r1r2r2=-3/5, not r1r2r3=-3/5.

- 5 years, 11 months ago

Could u please explain this to me in simple language as in just how do u apply it to quadratic equations and why is it even related to complex numbers? Thanks very much.

- 5 years, 11 months ago