Difference between revisions of "1965 AHSME Problems/Problem 10"

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\textbf{(C) }\ x < 3 \
 
\textbf{(C) }\ x < 3 \
 
\textbf{(D) }\ x > 3 \text{ and }x < - 2 \qquad  
 
\textbf{(D) }\ x > 3 \text{ and }x < - 2 \qquad  
\textbf{(E) }\ x > 3 \text{ and }x < - 2  </math>   
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\textbf{(E) }\ x > 3 \text{ or }x < - 2  </math>   
  
 
==Solution==
 
==Solution==

Latest revision as of 14:39, 13 November 2024

Problem 10

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

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

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 $\boxed{\textbf{(A)}}$.

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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
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