Difference between revisions of "1965 AHSME Problems/Problem 10"
(Created page with "== Problem 10== The statement <math>x^2 - x - 6 < 0</math> is equivalent to the statement: <math>\textbf{(A)}\ - 2 < x < 3 \qquad \textbf{(B) }\ x > - 2 \qquad \textbf{(C...") |
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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{ | + | \textbf{(E) }\ x > 3 \text{ or }x < - 2 </math> |
==Solution== | ==Solution== | ||
− | 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 | + | 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== | ||
+ | {{AHSME 40p box|year=1965|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 14:39, 13 November 2024
Problem 10
The statement is equivalent to the statement:
Solution
To solve this problem, we may begin by factoring as . This is an upward opening parabola, therefore the solutions to are between the roots of the equation. That means our solutions are all such that , or simply .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AHSME Problems and Solutions |
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