Difference between revisions of "1962 AHSME Problems/Problem 29"

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.

@@ Line 7: / Line 7: @@
 ==Solution==
-{{solution}}
+First, subtract 6 from both sides of the inequality,
+<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.

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Difference between revisions of "1962 AHSME Problems/Problem 29"

Revision as of 21:23, 4 June 2018

Problem

Solution