1965 AHSME Problems/Problem 10: Difference between revisions

Revision as of 09:59, 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 $\boxed{\textbf{(A)}}$ .

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

@@ Line 11: / Line 11: @@
 ==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 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>.
+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==

Page

Toolbox

Search

1965 AHSME Problems/Problem 10: Difference between revisions

Revision as of 09:59, 18 July 2024

Problem 10

Solution

See Also