Difference between revisions of "1980 AHSME Problems/Problem 6"
J314andrews (talk | contribs) (More detailed explanation of solution.) |
J314andrews (talk | contribs) m (Fixed parentheses) |
||
| Line 11: | Line 11: | ||
== Solution 2 (Graphing) == | == Solution 2 (Graphing) == | ||
| − | Graphing both <math>\sqrt{x}</math> and <math>2x</math> in Quadrant I of the coordinate plane, we see that the two curves intersect at <math>(\frac{1}{4}, \frac{1}{2})</math>, and <math>2x > \sqrt{x}</math> whenever <math>\boxed{(\textbf{A})\ x > \frac{1}{4}}</math>. | + | Graphing both <math>\sqrt{x}</math> and <math>2x</math> in Quadrant I of the coordinate plane, we see that the two curves intersect at <math>\left(\frac{1}{4}, \frac{1}{2}\right)</math>, and <math>2x > \sqrt{x}</math> whenever <math>\boxed{(\textbf{A})\ x > \frac{1}{4}}</math>. |
== See also == | == See also == | ||
{{AHSME box|year=1980|num-b=5|num-a=7}} | {{AHSME box|year=1980|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 02:14, 23 July 2025
Problem
A positive number 
satisfies the inequality 
if and only if
Solution 1
Since both sides of this inequality are positive, we may square both sides to get 
. Since is positive, we can divide both sides by to get 
. Dividing both sides by 
yields 
, or 
.
Solution 2 (Graphing)
Graphing both 
and 
in Quadrant I of the coordinate plane, we see that the two curves intersect at 
, and 
whenever .
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
