1965 AHSME Problems/Problem 10

Problem 10

The statement $x^2 - x - 6 < 0$ is equivalent to the statement:

$\textbf{(A)}\ - 2 < x < 3 \qquad \textbf{(B) }\ x > - 2 \qquad \textbf{(C) }\ x < 3 \\ \textbf{(D) }\ x > 3 \text{ and }x < - 2 \qquad \textbf{(E) }\ x > 3 \text{ and }x < - 2$

Solution

To solve this problem, we may begin by factoring $x^2-x-6$ as $(x-3)(x+2)$ . This is an upward opening parabola, therefore the solutions to $(x-3)(x+2) < 0$ are between the roots of the equation. That means our solutions are all $x$ such that $-2 < x < 3$ , or simply $\boxed{\textbf{(A)}}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1965 AHSME Problems/Problem 10

Problem 10

Solution

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