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Difference between revisions of "1962 AHSME Problems/Problem 29"
(→‎Solution)
 
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==Solution==
 
==Solution==
 
First, subtract 6 from both sides of the inequality,  
 
First, subtract 6 from both sides of the inequality,  
−
<math>2x^2 + x - 6 < 0</math>,
+
<math>2x^2 + x - 6 < 0</math>.
 
This is a parabola that opens upward when graphed, it has a positive leading coefficient. So any negative x values must be between its x-axis intersections, namely <math>x = -2, 1.5</math>. The answer is A.
 
This is a parabola that opens upward when graphed, it has a positive leading coefficient. So any negative x values must be between its x-axis intersections, namely <math>x = -2, 1.5</math>. The answer is A.

Latest revision as of 21:23, 4 June 2018

Problem

Which of the following sets of $x$-values satisfy the inequality $2x^2 + x < 6$?

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

$\textbf{(D)}\ \frac{3}2 < x < 2\qquad\textbf{(E)}\ x <-2$

Solution

First, subtract 6 from both sides of the inequality, $2x^2 + x - 6 < 0$. This is a parabola that opens upward when graphed, it has a positive leading coefficient. So any negative x values must be between its x-axis intersections, namely $x = -2, 1.5$. The answer is A.

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