# Difference between revisions of "1953 AHSME Problems/Problem 44"

## Problem

In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and obtains $8$ and $2$ for the roots. Another student makes a mistake only in the coefficient of the first degree term and find $-9$ and $-1$ for the roots. The correct equation was:

$\textbf{(A)}\ x^2-10x+9=0 \qquad \textbf{(B)}\ x^2+10x+9=0 \qquad \textbf{(C)}\ x^2-10x+16=0\\ \textbf{(D)}\ x^2-8x-9=0\qquad \textbf{(E)}\ \text{none of these}$

## Solution

Let $x^2+bx+c=0$ represent the correct equation. Since the coefficient of the $x^2$ term is $1$, the sum of the roots is $-b$, and the product of the roots is $c$.

If a student only misreads the constant term, he must have the correct sum of roots. Therefore, the sum of the roots is $8+2=10$, so $b=-10$. If a student only misreads the linear term, he must have the correct product of the roots. The product of the roots is $(-9)\cdot (-1) = 9$, so $c=9$. The correct equation is $\boxed{\textbf{(A) } x^2-10x+9=0}$.