Difference between revisions of "1971 AHSME Problems/Problem 8"
Coolmath34 (talk | contribs) (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> | ||
− | The graph of this inequality is a parabola facing upwards, so the | + | 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 | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME 35p box|year=1971|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:16, 1 August 2024
Problem
The solution set of is the set of all values of
such that
Solution
We are solving the inequality This can be factored as
The graph of this inequality is a parabola facing upwards, so the interval between the roots satisfies the equation. This interval, , is answer
.
-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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.