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

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==Solution==
 
==Solution==
  
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To solve this problem, we may begin by factoring <math>x^2-x-6</math> as <math>(x-3)(x+2)</math>. This is an upward opening parabola, therefore the solutions to <math>(x-3)(x+2) < 0</math> are inbetween the roots of the equation. That means our solutions are all <math>x</math> such that <math>-2 < x < 3</math>, or simply <math>\boxed{\textbf{(A)}}</math>.
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To solve this problem, we may begin by factoring <math>x^2-x-6</math> as <math>(x-3)(x+2)</math>. This is an upward opening parabola, therefore the solutions to <math>(x-3)(x+2) < 0</math> are between the roots of the equation. That means our solutions are all <math>x</math> such that <math>-2 < x < 3</math>, or simply <math>\boxed{\textbf{(A)}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1965|num-b=9|num-a=11}}
 
{{AHSME 40p box|year=1965|num-b=9|num-a=11}}
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{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 17:03, 18 July 2024

Problem 10

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

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

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

See Also

1965 AHSC (Problems • Answer Key • Resources)
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
All AHSME Problems and Solutions

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