Difference between revisions of "1965 AHSME Problems/Problem 10"
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==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== | ==See Also== |
Revision as of 09:59, 18 July 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 |