Vieta's Formula

Consider a quadratic equation with complex coefficients and roots r1r_1, r2r_2 : a2(xr1)(xr2)=a2x2+a1x+a0. a_2(x-r_1)(x-r_2) = a_2x^2+a_1x+a_0. By comparing coefficients, we can see that r1+r2=a1a2,r1r2=a0a2. 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 nn, we have

Vieta's Formula. Let P(x)=anxn+an1xn1++a0P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 be a polynomial with complex coefficients and degree nn, having complex roots rn,rn1,,r1r_n, r_{n-1}, \cdots, r_1. Then for any integer 0kn0 \leq k \leq n:

1i1<i2<<iknri1ri2rik=(1)kankan. \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 (xri) \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:

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

Solution: Any 3 numbers (a,b,c) (a,b,c) can be considered as the roots of the monic cubic polynomial P(x)=(xa)(xb)(xc)P(x) = (x-a)(x-b)(x-c) . Applying Vieta's Formula to the polynomial, we get P(x)=x3P(x)=x^3. Hence, the only root of P(x)P(x) is x=0x=0 (with multiplicity 3), which implies that (0,0,0)(0,0,0) is the only triple of complex numbers that satisfies the system of equations.

 

2. Let r1,r2r_1, r_2 and r3r_3 be the roots of the polynomial 5x311x2+7x+35x^3 -11x^2+7x+3. Evaluate r1(1+r2+r3)+r2(1+r3+r1)+r3(1+r1+r2).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 r1+r2+r3+2(r1r2+r2r3+r3r1)r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1) . By Vieta's Formula, we know that

r1+r2+r3=115=115,r1r2+r2r3+r3r1=75,r1r2r3=35, 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 r1+r2+r3+2(r1r2+r2r3+r3r1)=115+2×75=255=5r_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 r1,r2r_1, r_2 and r3r_3 be the roots of the polynomial 5x311x2+7x+35x^3 -11x^2+7x+3. Evaluate r13+r23+r33.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

r13+r23+r33=3r1r2r3+(r1+r2+r3)[r12+r22+r32(r1r2+r2r3+r3r1)].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 r12+r22+r32r_1^2+r_2^2+r_3^2. Again, from factorization, we have that

r12+r22+r32=(r1+r2+r3)22(r1r2+r2r3+r3r1), 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

r13+r23+r33=3r1r2r3+(r1+r2+r3)[r12+r22+r32(r1r2+r2r3+r3r1)]=3r1r2r3+(r1+r2+r3)[(r1+r2+r3)23(r1r2+r2r3+r3r1)]=3×(35)+(115)[(115)23×75]=49125. \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 aa for which the equation

16x4ax3+(2a+17)x2ax+16=0 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 aa satisfies the requirements of the problem and that xx, qxqx, q2xq^2x, q3xq^3x are the roots of the given equation. Then x0x \neq 0 and we may assume that q>1|q|>1, so that x<qx<q2x<q3x|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, α\alpha is a root, then its reciprocal (which is α1=1α\alpha ^{-1} = \frac{1}{\alpha}) will also be a root. Hence, 1x=q3x, \frac{1}{x} = q^3 x, so q=x23 q = x^{-\frac{2}{3} } and the roots are x,x13,x13,x1x, x^{\frac{1}{3}}, x^{\frac{-1}{3}}, x^{-1}.

Now, by Vieta's formula we have x+x13+x13+x1=a16x+x^{\frac{1}{3}}+x^{\frac{-1}{3}}+x^{-1}=\frac{a}{16} and x43+x23+1+1+x23+x43=2a+1716x^{\frac{4}{3}}+x^{\frac{2}{3}}+1 + 1+x^{\frac{-2}{3}}+x^{\frac{-4}{3}}=\frac{2a+17}{16}. On setting z=x13+x13z =x^{\frac{1}{3}}+x^{\frac{-1}{3}} these equations become

z32z=a16,(z22)2+z22=2a+1716. z^3-2z=\frac{a}{16}, (z^2-2)^2+z^2-2=\frac{2a+17}{16}.

Substituting a=16(z3z2)a=16(z^3-z^2) in the second equation leads to z42z33z2+4z+1516=0z^4-2z^3-3z^2+4z+\frac{15}{16} = 0. We observe that this polynomial factors as (z+32)(z52)(z2z14)(z+\frac{3}{2})(z-\frac{5}{2})(z^2-z-\frac{1}{4}). Since z=x13+x132|z|=|x^{\frac{1}{3}}+x^{\frac{-1}{3}}|\geq 2, the only viable value is z=52z=\frac{5}{2}. Consequently a=170a=170

Rearranging the equation, we get 16(x2+1x2)170(x+1x)+357=016 (x^2 + \frac{1}{x^2}) - 170(x + \frac{1}{x}) + 357 = 0. To simplify it, we can call y=x+1xy = x + \frac{1}{x} and therefore y2=x2+2+1x2y^2 = x^2 + 2 + \frac{1}{x^2}, thus getting a new form 16(y22)170y+357=016y2170y+325=016 (y^2 - 2) - 170y + 357 = 0 \Rightarrow 16y^2 - 170y +325 = 0, whose roots are 52\frac{5}{2} and 658\frac{65}{8}. We have to plug the two back into y=x+1xy = x + \frac{1}{x}, leading us to two more quadratics, getting finally 18,12,2,8\boxed{\dfrac{1}{8}, \dfrac{1}{2}, 2, 8}.


Test Yourself

  1. If α,β \alpha, \beta are the roots to the equation x2+2x+3=0 x^2 +2x+3 = 0 , what is the quadratic equation whose roots are (α1α)2 (\alpha - \frac{1}{\alpha} )^2 and (β1β)2 ( \beta - \frac{1}{\beta} ) ^ 2 ?

  2. ARML 2012, Team Problems, #6 The zeros of f(x)=x6+2x5+3x4+5x3+8x2+13x+21f(x) = x^6 + 2x^5 + 3x^4 + 5x^3 + 8x^2 + 13x + 21 are distinct complex numbers. Compute the average value of A+BC+DEFA + BC + DEF over all possible permutations (A,B,C,D,E,F)(A, B, C, D, E, F) of these six numbers.

  3. (Created by Mark Hennings. This is HARD) Show that j=1ncot2(jπ2n+1)  =  13n(2n1) \sum_{j=1}^n \cot^2\Big(\tfrac{j \pi}{2n+1}\Big) \; = \; \tfrac13n(2n-1) for any integer n1n \ge 1. What is j=1ncot4(jπ2n+1)? \sum_{j=1}^n \cot^4\Big(\tfrac{j \pi}{2n+1}\Big) \qquad \mbox{?}

Note by Calvin Lin
5 years, 7 months ago

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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.

Tan Yong Boon - 4 years, 12 months ago

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

George Darroch - 5 years ago

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