## Problem 2

Find all real roots of the quartic $\mathrm{x}^{4}-(2 \mathrm{~N}+1) \mathrm{x}^{2}-x+\mathrm{N}^{2}+N-1=0$ correct to 4 decimal places, where $\mathrm{N}=$ $10^{10}$.


## Solution

Answer: 99999.9984 and 100000.0016 .

We can write the equation as $\left(x^{2}-N-1 / 2\right)^{2}=x+5 / 4$. For $x329^{2}-10^{5}-2>0$. So indeed one root lies between $10^{5}-0.00165$ and $10^{5}-0.00155$ and the other root lies between $10^{5}+0.00155$ and $10^{5}+0.00165$.

## Descartes' rule of signs.

This states that if the number of sign changes in the coefficients of the polynomial is $d$, then the number of positive roots is $\mathrm{d}$ or $\mathrm{d}$ less an even number. So, for example, if the polynomial is $\mathrm{x}^{5}+14.3 \mathrm{x}^{4}-34 \mathrm{x}^{2}-\mathrm{x}+$ 3.2, then there are two sign changes ( +14.3 to -34 and -1 to 3.2 ), so there are either 0 or 2 positive roots. Note that we ignore zero coefficients. If $r$ is a positive root of $p(-x)=0$, then $-r$ is a negative root of $p(x)=0$. So if we substitute - $\mathrm{x}$ for $\mathrm{x}$ in the polynomial and the number of sign changes is then $\mathrm{d}^{\prime}$, then we can conclude that the number of negative roots of the polynomial is either $\mathrm{d}^{\prime}$ or $\mathrm{d}^{\prime}$ less an even number. With the example above we get $-x^{5}+14.3 x^{4}-34 x^{2}+x+3.2$, which has 3 sign changes. So the polynomial has 1 or 3 negative roots.

The proof is not difficult. The key idea is to show that if $k$ is a positive root, so that $p(x)=(x-k) q(x)$, then (1) $p(x)$ has at least one more sign change than $q(x)$, and (2) the difference between the number of sign changes is odd (note that the signs of the constant coefficients of $p(x)$ and $q(x)$ are different).

