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

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 43	Followed by Problem 45
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1953 AHSME Problems/Problem 44

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