# Vieta's Formula

This week, we learn about Vieta's Formula, which allows us to relate the roots of a polynomial to its coefficients.

How would you use Vieta's Formula to solve the following?

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. Share a problem which requires understanding of Vieta’s Formula.

Note by Calvin Lin
6 years, 2 months ago

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Perhaps a slightly simpler solution for part 1.

Let the required quadratic equation be: $x^2 - Ax + B = 0$, then we have:

\begin{aligned} A & = \left(\alpha - \frac{1}{\alpha} \right)^2 + \left(\beta - \frac{1}{\beta} \right)^2 = \alpha^2 - 2 + \frac{1}{\alpha^2} + \beta^2 - 2 + \frac{1}{\beta^2} = \alpha^2 + \beta^2 - 4 + \frac{1}{\alpha^2} + \frac{1}{\beta^2} \\ & = (\alpha + \beta)^2 - 2 \alpha \beta - 4 + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)^2 - \frac{2}{ \alpha \beta} = (-2)^2 - 2(3) - 4 + \left(\frac{\alpha + \beta}{\alpha \beta}\right)^2 - \frac{2}{3} \\ & = 4 - 6 - 4 + \frac{4}{9} - \frac{2}{3} = -\frac{56}{9} \\ B & = \left(\alpha - \frac{1}{\alpha} \right)^2 \left(\beta - \frac{1}{\beta} \right)^2 = \left(\alpha\beta - \frac{\alpha}{\beta} - \frac{\beta}{\alpha} + \frac{1}{\alpha \beta} \right)^2 = \left(3 - \frac{\alpha^2 + \beta^2} {\beta \alpha} + \frac{1}{3} \right)^2 \\ & = \left(3 - \frac{(-2)^2 - 2(3)} {3} + \frac{1}{3} \right)^2 = \left(3 + \frac{2} {3} + \frac{1}{3} \right)^2 = 4^2 = 16 \end{aligned}

$\Rightarrow x^2 - Ax + B = 0 \\ x^2 + \dfrac{56}{9}x + 16 = 0 \\ \boxed{9x^2 + 56x + 16 = 0}$

- 4 years, 6 months ago

I don't really get vieta's theorem. I know that it us used to solve quadratics and that I am able to use it. The wiki shows some very complicated definition with complex numbers involved. Can someone please help clarify my doubts? Thanks very much.

- 5 years, 4 months ago

*it is

- 5 years, 4 months ago

The product of two of the four zeros of the quadrtic equation

x^4 - 18x^3 + kx^2 + 200x - 1984 = 0 is -32 . then find k

- 6 years, 2 months ago

Answer is 86... By some algebraic manipulation..

- 6 years, 2 months ago

what is the next two numbers for the series: 0,2,24......?

- 6 years, 2 months ago

i think so answer to this question should be 246

- 6 years, 2 months ago

what is the next two numbers

Tushar, we'd have to find the next one ($2468$).

- 6 years, 2 months ago

This question has infinite answers. I'd answer $\pi^{100}, e^{200}$. 246 and 2468 could be nice guesses, but we cannot state this is the absolute true answer.

EDIT: Also, not related to Vieta's Formula.

- 6 years, 2 months ago

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.

- 6 years, 2 months ago

First, there are ${6 \choose 3} {3 \choose 2} = 60$ permutations. So, this is equal to $10(A + B + C + D + E + F) + 4(AB + BC + CD + DE + EF + ...) + 3(ABC + ABD + ABE + ...)$. Using Vieta's formula, this is equal to $10(-2) +4(3) + 3(-5) = -23$. Taking the average value, this is $\frac{-23}{60}$.

- 6 years, 2 months ago

(GCDC)
If $\displaystyle \sum_{a=0}^5 a \times x^a = \displaystyle \prod_{b=1}^{5} (x-x_b)$, evaluate $\displaystyle \sum_{b=1}^5 -x_b \: ^5$.

EDIT: The answer is $113$.

- 6 years, 2 months ago

The desired polynomial is $5x^5 + 4x^4 + 3x^3 + 2x^2 + x$. We wish to find the sum of the fifth powers of the roots $a, b, c, d, e$. I am confident that a solution can be found using the expansion of $(a + b + c + d + e)^5$ and further and further factorization, and chasing the solution further in this way is trivial. I'll edit the post if I think of more clever way to find the solution.

- 6 years, 2 months ago

You can first find the values of $a^2 + b^2 + c^2 + d^2 + e^2$ and $a^3 + b^3 + c^3 + d^3 + e^3$. Then find their product. It's much less work compared to actually expanding $(a+b+c+d+e)^5$. But still look ridiculously long compared to Newton's Sums.

- 6 years, 2 months ago

For $2 \leq n \leq 3$, the absolute value of $\displaystyle \sum_{b=1}^n x_b \: ^n$ still looks very easy and nice to do by Vieta's Formulae.

For $n \geq 4$, things begin to get "olympic", thus, as Pi Han Goh stated, making Newton's Sums use much preferable.

- 6 years, 2 months ago

For the given problem ,we use Vieta's Formula and get :

$\alpha + \beta = - \dfrac{b}{a} = -2$ and $\alpha \cdot \beta = \dfrac{c}{a} = 3$

[Now - considering the case that Calvin sir wanted it to be $(\alpha - \dfrac{1}{\alpha} )^2$]

Let the new equation be -

$Ax^2 + Bx + C = 0$

Thus by again applying * Vieta's Formula * ,

We get the following relations between the assumed coefficients and the* required *roots :

$-\dfrac{B}{A} = \left(\alpha - \dfrac{1}{\alpha} \right)^2 + \left(\beta - \dfrac{1}{\beta} \right)^2$

$-\dfrac{B}{A} = \alpha ^2 - 2 + \left(\dfrac{1}{\alpha} \right)^2 + \beta^2 - 2 + \left(\dfrac{1}{\beta} \right)^2$

To obtain the value of LHS , we convert RHS in the form of $\alpha + \beta$ and $\alpha \cdot \beta$.

$-\dfrac{B}{A} = ( \alpha + \beta )^2 - 2 \alpha \beta + \left[ \dfrac{1}{\alpha} + \dfrac{1}{\beta} \right]^2 - \dfrac{2}{\alpha\beta} - 4$

$-\dfrac{B}{A} = ( \alpha + \beta )^2 - 2 \alpha \beta + \left[ \dfrac{\alpha + \beta}{ \alpha \beta} \right]^2 - \dfrac{2}{\alpha\beta} - 4$

Thus by putting the values we obtained initially , we have -

$-\dfrac{B}{A} = ( -2)^2 - 2 \cdot 3 + \left[ \dfrac{-2}{ 3} \right]^2 - \dfrac{2}{3} - 4$

This gives us : [CAUTION : I cancelled the negative signs of both sides]

$\dfrac{B}{A} =\boxed{\dfrac{56}{9}}$

Similarly we get :

$\dfrac{C}{A} = \left(\alpha - \dfrac{1}{\alpha} \right)^2 \times \left(\beta - \dfrac{1}{\beta} \right)^2$

$\dfrac{C}{A} = (\alpha \beta)^2 + \left[ \dfrac{1}{\alpha \beta} \right]^2 - 2(\alpha^2 + \beta^2) + \dfrac{(\alpha^2 + \beta^2 )^2-2 \alpha^2 \beta^2 }{\alpha^2 \beta^2} - 2\left( \dfrac{\alpha^2 + \beta^2 }{\alpha^2 \beta^2 } \right) + 4$

Thus , we put the values of $\alpha + \beta$ , $\alpha \beta$ and $\alpha^2 + \beta^2$ [Refer above]

We get - $\dfrac{C}{A} = \dfrac{16}{1} = \boxed{\dfrac{144}{9}}$

Thus by comparing the values of $A$ , $B$ and $C$ -

We get $A = 9$ $B = 56$ $C = 144$

And hence , the required quadratic equation is -

$\boxed{ 9x^2 + 56x + 144 = 0 }$

- 6 years, 2 months ago

An easier way is to first find the quadratic equation which has roots $\alpha - \frac {1}{\alpha}$ and $\beta - \frac {1}{\beta}$

By Vieta's formula, we have $\alpha + \beta = -2$ and $\alpha \beta = 3$, then $\alpha^2 + \beta^2 = ( \alpha + \beta)^2 - 2 \alpha \beta = (-2)^2 - 2(3) = -2$

Now consider a quadratic equation $f(x)=ax^2+bx+c$ with integers $a>0,b,c$ which has roots $\alpha - \frac {1}{\alpha}$ and $\beta - \frac {1}{\beta}$

The sum of roots is $\alpha - \frac {1}{\alpha} + \beta - \frac {1}{\beta} = \alpha + \beta - ( \frac {1}{\alpha} + \frac {1}{\beta} ) = ( \alpha + \beta ) - ( \frac { \alpha + \beta }{\alpha \beta } ) = ( \alpha + \beta ) (1 - \frac {1}{\alpha \beta} ) = (-2)(1 - \frac {1}{3} ) = - \frac {4}{3}$

And the product of roots is

$(\alpha - \frac {1}{\alpha})(\beta - \frac {1}{\beta}) = \alpha \beta - ( \frac {\alpha}{\beta} + \frac {\beta}{\alpha} ) + \frac {1}{\alpha \beta} = \alpha \beta - \frac {\alpha^2 + \beta^2 - 1}{\alpha \beta} = 3 - \frac {-2-1}{3} = 4$

Then $f(x) = 3( x^2 - (- \frac {4}{3} ) x + 4 ) = 3x^2 +4x + 12$ has roots $\alpha - \frac {1}{\alpha}$ and $\beta - \frac {1}{\beta}$

Thus $f(\sqrt{x}) = 0$ has roots $(\alpha - \frac {1}{\alpha})^2$ and $(\beta - \frac {1}{\beta})^2$

$\Rightarrow 3 ( \sqrt{x} )^2 + 4 \sqrt{x} + 12 = 0$

$\Rightarrow 3x + 12 = -4 \sqrt{x}$

$\Rightarrow (3x+12)^2 = 16x$

$\Rightarrow 9x^2 + 56x + 144 = 0$ has roots $(\alpha - \frac {1}{\alpha})^2$ and $(\beta - \frac {1}{\beta})^2$

- 6 years, 2 months ago

Good approach!

I like how you use $f ( \sqrt{x})$ to square the roots.

Staff - 6 years, 2 months ago

Hey I just wanted to ask, in the 4th and final question in Vieta's formula, how did you find out the roots by just seeing the equation ? Please do explain.

- 6 years, 2 months ago

It becomes a reciprocal polynomial, which is very easy to solve. Those polynomials have roots which multiplied together give 1. Generally, $x_1$ and $\frac{1}{x_1}$.

- 6 years, 2 months ago

In addition, if a polynomial is of the form $P(x) = Ax^n + Bx^{n - 1} + \cdots\ + Yx + Z$ and another polynomial is of the form $Q(x) = Zx^n + Yx^{n-1} \cdots + Bx + A$, then the roots of Q are the reciprocals of the roots of P.

- 6 years, 2 months ago

- 6 years, 2 months ago

Calvin explained it a little on his last worked example.

Note that for this polynomial (which Calvin didn't write, but is implicit) $16x^4-170x^3+357x^2-170x+16x=0 \;$ the coefficients are symmetrical (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, let's say, $\alpha$ is a root, its reciprocal (which is $\alpha ^{-1} = \frac{1}{\alpha}$) will also be a root.

Because the independent coefficient ($x^0$) is not zero, we can assure that no root equals zero. Therefore we can divide the whole equation by a power of $x$, which here would be $x^2$. (Can you figure out why by yourself?)

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 $16 (y^2 - 2) - 170y + 357 = 0 \Rightarrow 16y^2 - 170y +325 = 0$, a quadratic equation easily solvable by any known formula.

Attention now: solving $16y^2 - 170y +325 = 0$ gives us $\frac{5}{2}$ and $\frac{65}{8}$, the values of $y$, not $x$. We have to plug the two back into $y = x + \frac{1}{x}$, leading us to two more quadratics, getting finally $\boxed{\dfrac{1}{8}, \dfrac{1}{2}, 2, 8}$.

EDIT: Sorry for being too long. Hope you understood.

- 6 years, 2 months ago

Staff - 6 years, 2 months ago

Woohoo, a personal accomplishment achieved! \o/

I just felt it'd be much better to explain the full story right from the start to one who isn't familiar to reciprocal polynomials. We have three more other cases, which I really want to work with, but now is not the adequate time (since we are only analising Vieta's Formulae).

- 6 years, 2 months ago

Thank you for explaining, now may I ask, can we proof for a general symmetric equation without solving it, that it has symmetric roots ?

- 6 years, 2 months ago

Let's consider a polynomial like the ones we worked with, $P(x) = \alpha x^{2n} + \beta x^{2n-1} + \dots + \omega + \dots + \beta x + \alpha$. The condition for it to be an even degree polynomial is that $\alpha \neq 0$. Therefore the product of its roots (which is $\frac{\alpha}{\alpha} = 1$, but let's pretend we don't know that) is not zero, so none of the roots equal zero, allowing us to do $\dfrac{P(x)}{x^n} = 0$.

Its form now is $P'(x) = \alpha (x^n + x^{-n}) + \beta (x^{n-1} + x^{1-n}) + \dots + \omega$. Making more changes because $x \neq 0$, we can express it as a function of $y = x + x^{-1}$, becoming $Q(y) = \omega + \psi y + \chi (y^2 - 2) + \dots \;$.

This new polynomial has roots $y = A, \; B, \; \dots, \; Z$, let's say. After finding them, we plug them back into the equation $y = x + x^{-1}$. Let's pick $A$ for this example. Once again because $x \neq 0$ we can multiply the $y$ equation by $x$, finding a quadratic $x^2 - Ax + 1 = 0$. Note that, for any root of $Q(x)$, the product of this quadratic's roots will always be 1, guaranteeing us that if $P(x_0) = 0$, then $P(\frac{1}{x_0}) = 0$.

- 6 years, 2 months ago

I'd highly recommend you work through the proof yourself using the example Guilherme provided. The general case is exactly the same. Just apply the method he used to the polynomial $p(x) = a_nx^n + ... + a_{\frac{n}{2} + 1}x^{\frac{n}{2} + 1} + a_{\frac{n}{2}}x^{\frac{n}{2}} + a_{\frac{n}{2} + 1}x^{\frac{n}{2} - 1} + ... + a_nx^0$ with $a_i \in \mathbb{C}, a_n \neq 0, n$ an even positive integer (this is the general symmetric polynomial). It may seem intimidating, but just work through it in the exact same way Guilherme did in his example. Remember, also, that you have it easier: he went through and calculated the roots. All you need to do is show that if $r$ is a root, so is $\frac{1}{r}$. You don't care at all what those roots actually are, so don't do any of the substitutions with $y$. It may be useful to note that $r + \frac{1}{r} = \frac{1}{r} + \frac{1}{\frac{1}{r}}$. Good luck :)

- 6 years, 2 months ago

Do you really mean for $\beta-\beta^{-1}$ to be squared and $\alpha-\alpha^{-1}$ not?

- 6 years, 2 months ago

Edited, thanks!

Staff - 6 years, 2 months ago

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

- 6 years, 2 months ago

I like this one. The idea is to show that $\prod_{j=1}^n \left( x^2 - \cot^2( \frac{j\pi}{2n+1} ) \right) = \frac1{2n+1} {\rm Im} (x+i)^{2n+1}$ by comparing roots; then compare coefficients of $x^{2n-2}$.

- 6 years, 2 months ago

Another way to look at it is to (courtesy of de Moivre's Theorem) find the degree $n$ polynomial $f_n(X)$ such that $\sin(2n+1)x \; = \; \sin^{2n+1}x f_n(\cot^2x)$ and identify the roots of $f_n(X)$.

What is cool is that you can use these identities to obtain low-level proofs of the series $\sum_{n=1}^\infty \frac{1}{n^2} \; = \; \tfrac16\pi^2 \qquad \sum_{n=1}^\infty \frac{1}{n^4} \; = \; \tfrac{1}{90}\pi^4$ since you can use them to obtain error bounds on the partial sums.

- 6 years, 2 months ago

Let $m, n > 1$ and $m, n \in \mathbb{Z}$. Suppose that the product of the solutions for x of the equation $8(\log_n x)(\log_m x) - 7\log_n x - 6 \log_m x - 2013 = 0$ is the smallest possible integer. Compute $m + n$

Source: AMC

- 6 years, 2 months ago

Hint: Write $\log_a b$ as $\frac{log_b}{log_a}$.

..

Solution:

Using the hint, we have $(\frac{8}{(\log n)(\log m)}) (\log x)^2 - (\frac {7}{\log n} + \frac {6}{\log m})\log x - 2013 = 0$. This is quadratic in $\log x$. Using vieta's formulas, we have that the sum of the roots $a_1 = \log x_1, a_2 = \log x_2$ is equal to $\frac{\frac {7}{\log n} + \frac {6}{\log m}}{\frac{8}{\log n \log m}} = \frac {7 \log m + 6 \log n}{8}$.

Using log properties, this expression is equal to $\frac {\log m^7 n^6}{8} = \log \sqrt[8] {m^7 n^6}$ Because $a_1 + a_2 = \log x_1 + \log x_2 = \log x_1 x_2$, the product of the solutions for $x$ is equal to $\sqrt[8]{m^7 n^6}$. To minimize this as an integer and $m , n > 1$, we have $m = 2^a, n = 2^b$ and find the smallest multiple of $8$ which can be written as $7a + 6b$ for $a, b \geq 1$. Since $6 \equiv -2 \pmod 8$ and $7 \equiv -1 \pmod 8$ it is easy to find a solution in $a = 2, b = 3$. Thus, $x_1x_2$ is minimized at $m = 8, n = 4$, and $m + n = 12$.

- 6 years, 2 months ago

Hard problem, nice solution!

- 6 years, 2 months ago