0
$\begingroup$

Consider the quadratic equation $a_2x^2+a_1x+a_0=0$

If $x_1$ and $x_2$ are the roots of the quadratic then its Vieta's formulas are : $\left\{ \begin{align} x_1 + x_2 &= -\dfrac{a_1}{a_2}\\ x_1 \cdot x_2 &= \dfrac{a_0}{a_2} \end{align} \right.$

In a similar manner if $x_1,x_2,x_3$ are the three roots of $a_3x^3+a_2x^2+a_1x+a_0=0$ then its Vieta's formulas are : $$\left\{ \begin{align} x_1 + x_2 + x_3 &= -\dfrac{a_2}{a_3}\\ x_1 \cdot x_2 \cdot x_3 &= -\dfrac{a_0}{a_3}\\ x_1x_2+x_1x_3+x_2x_3 &= \dfrac{a_1}{a_3} \end{align} \right.$$

Should I do one more? If you consider $a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=0$ and $x_1,x_2,x_3,x_4$ are the 4 roots of the equation then it's possible to give such 4 formulas, you may call those Vieta's foemulas but my question is what's the use of it?

For any n degree polynomial there exists such n formulas. Should I remember those all the times to solve some handcrafted problems?

And how many of those you do remember?

$\endgroup$
1
$\begingroup$

An infinity.

The formulas are pretty regular. You form the sums of all distinct products of $n-k$ coefficients and equate them to the ratio of the $k^{th}$ coefficient and the leading one.

From scratch, for $n=4$,

$$\begin{cases}x_1+x_2+x_3+x_4=-\frac{a_3}{a_4}, \\x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=\frac{a_2}{a_4}, \\x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=-\frac{a_1}{a_4}, \\x_1x_2x_3x_4=\frac{a_0}{a_4}.\end{cases}$$

The number of terms follows the Binomial distribution, $(1),4,6,4,1$.

$\endgroup$
  • $\begingroup$ Would you please write those formulas for $n=6$ and give me a hint about why the number of terms follow the Binomial distribution? $\endgroup$ – ARahman Oct 11 '18 at 16:32
  • $\begingroup$ @ARahman: nope, your job. Think of combinations. $\endgroup$ – Yves Daoust Oct 11 '18 at 17:58
0
$\begingroup$

First, it's easier if you assume the leading coefficient (i.e. the coefficient of $x^n)$ is $1,$ which you can always arrange in practice by factoring out the leading coefficient. Then the formulas follow a simple pattern, and you only have to remember the pattern.

$-\left(\text{sum of products-one-at-a-time of the roots}\right)=$ coefficient of $x^{n-1}$

$+\left(\text{sum of products-two-at-a-time of the roots}\right)=$ coefficient of $x^{n-2}$

$-\left(\text{sum of products-three-at-a-time of the roots}\right)=$ coefficient of $x^{n-3}$

$$ \cdots $$

$(-1)^{n-1}\cdot\left(\text{sum of products-}(n-1)\text{-at-a-time of the roots}\right)=$ coefficient of $x^{n-(n-1)} = x^1$

$(-1)^{n}\cdot\left(\text{sum of products-}n\text{-at-a-time of the roots}\right)=$ coefficient of $x^{n-n} = x^0$ (constant term)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.