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Difference between revisions of "1971 AHSME Problems/Problem 8"

(Created page with "== Problem == The solution set of <math>6x^2+5x<4</math> is the set of all values of <math>x</math> such that <math>\textbf{(A) }\textstyle -2<x<1\qquad \textbf{(B) }-\frac...")
 
m (see also box, made stated answer the correct answer B, not C (as derived), fixed other inaccuracies)
 
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<cmath>(2x-1)(3x+4) < 0</cmath>
 
<cmath>(2x-1)(3x+4) < 0</cmath>
  
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The graph of this inequality is a parabola facing upwards, so the area between the roots satisfies the equation. The solution is <math>x \in [-\frac{4}{3}, \frac{1}{2}]</math> and the answer is <math>\textbf{(C)}.</math>
+
The graph of this inequality is a parabola facing upwards, so the [[interval]] between the roots satisfies the equation. This interval, <math>(-\frac{4}{3}, \frac{1}{2})</math>, is answer <math>\boxed{\textbf{(B)}}</math>.
  
 
-edited by coolmath34
 
-edited by coolmath34
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== See Also ==
 +
{{AHSME 35p box|year=1971|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 11:16, 1 August 2024

Problem

The solution set of $6x^2+5x<4$ is the set of all values of $x$ such that

$\textbf{(A) }\textstyle -2<x<1\qquad \textbf{(B) }-\frac{4}{3}<x<\frac{1}{2}\qquad \textbf{(C) }-\frac{1}{2}<x<\frac{4}{3}\qquad \\ \textbf{(D) }x<\textstyle\frac{1}{2}\text{ or }x>-\frac{4}{3}\qquad \textbf{(E) }x<-\frac{4}{3}\text{ or }x>\frac{1}{2}$

Solution

We are solving the inequality $6x^2 + 5x - 4 < 0.$ This can be factored as \[(2x-1)(3x+4) < 0\]

The graph of this inequality is a parabola facing upwards, so the interval between the roots satisfies the equation. This interval, $(-\frac{4}{3}, \frac{1}{2})$, is answer $\boxed{\textbf{(B)}}$.

-edited by coolmath34

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
All AHSME Problems and Solutions

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