# Difference between revisions of "2018 AMC 12A Problems/Problem 21"

## Problem

Which of the following polynomials has the greatest real root?

$\textbf{(A) } x^{19}+2018x^{11}+1 \qquad \textbf{(B) } x^{17}+2018x^{11}+1 \qquad \textbf{(C) } x^{19}+2018x^{13}+1 \qquad \textbf{(D) } x^{17}+2018x^{13}+1 \qquad \textbf{(E) } 2019x+2018$

## Solutions

### Solution 1

We can see that our real solution has to lie in the open interval $(-1,0)$. From there, note that $x^a < x^b$ if $a$, $b$ are odd positive integers if $a, so hence it can only either be B or E(as all of the other polynomials will be larger than the polynomial B). E gives the solution $x=-\frac{2018}{2019}$. We can approximate the root for B by using $x=-\frac 12$. $$(- \frac {1}{2}) ^{17} - \frac{2018}{2048} + 1 \approx 0$$ therefore the root for B is approximately $-\frac 12$. The answer is $\fbox{B}$. (cpma213)

### Solution 2 (Calculus version of solution 1)

Note that $a(-1)=b(-1)=c(-1)=d(-1) < 0$ and $a(0)=b(0)=c(0)=d(0) > 0$. Calculating the definite integral for each function on the interval $[-1,0]$, we see that $B(x)\rvert^{0}_{-1}$ gives the most negative value. To maximize our real root, we want to maximize the area of the curve under the x-axis, which means we want our integral to be as negative as possible and thus the answer is $\fbox{B}$.

### Solution 3 (Alternate Calculus Version)

Newton's Method is used to approximate the zero $x_{1}$ of any real valued function given an estimation for the root $x_{0}$: $x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}\,.$ After looking at all the options, $x_{0}=-1$ gives a reasonable estimate. For options A to D, $f(-1) = -2018$ and the estimation becomes $x_{1}=-1+{\frac {2018}{f'(-1)}}\,.$ Thus we need to minimize the derivative, giving us B. Now after comparing B and E through Newton's method, we see that B has the higher root, so the answer is $\fbox{B}$. (Qcumber)

### Solution 4

Let the real solution to $B$ be $a.$ It is easy to see that when $a$ is plugged in to $A,$ since $-1 < a < 0,$ $a^{19} < a^{17}$ thus making the real solution to $A$ more "negative", or smaller than $B.$ Similarly we can assert that $D > C.$ Now to compare $B$ and $D,$ we can use the same method to what we used before to compare $B$ to $A,$ in which it is easy to see that the smaller exponent $(11)$ "wins". Now, the only thing left is for us to compare $B$ and $E.$ Plugging $\frac{-2018}{2019}$ (or the solution to $E$) into $B$ we obtain $\frac{(-2018)^{17}}{2019^{17}} + 2018\frac{(-2018)^{11}}{2019^{11}} + 1,$ which is intuitively close to $-1 - 2018 + 1 = -2018,$ much smaller than the solution the required $0.$ (For a more rigorous proof, one can note that $(\frac{2018}{2019})^{17}$ and $(\frac{2018}{2019}^{11})$ are both much greater than $(\frac{2018}{2019})^{2019} \approx \frac{1}{e},$ by the limit definition of $e.$ Since $- \frac{1}{e} - 2018 \cdot \frac{1}{e} + 1$ is still much smaller the required $0$ for the solution to $B$ to be a solution, our answer is $\boxed{B}.$ -fidgetboss_4000

~ dolphin7