# 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 $\boxed{\textbf{(A) } x^2-10x+9=0}$.

## See Also

 1953 AHSC (Problems • Answer Key • Resources) Preceded byProblem 43 Followed byProblem 45 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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