Difference between revisions of "1968 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | <math>\fbox{D}</math> |
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| + | <math>\frac{a+b}{2}=2\cdot\sqrt{ab}</math> | ||
| + | |||
| + | <math>\frac{a}{b} +1=4\cdot\sqrt{\frac{a}{b}}</math> | ||
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| + | setting <math>x=\sqrt{\frac{a}{b}}</math> we get a quadratic equation is<math>x^2+1=4x</math> with solutions <math>x=\frac{4\pm \sqrt{16-4}}{2}</math> | ||
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| + | <math>x^2=\frac{a}{b}=(4+3)+4\sqrt{3}=13.8=14</math>. | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=27|num-a=29}} | + | {{AHSME 35p box|year=1968|num-b=27|num-a=29}} |
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 01:53, 16 August 2023
Problem
If the arithmetic mean of 
and 
is double their geometric mean, with 
, then a possible value for the ratio 
, to the nearest integer, is:
Solution
setting 
we get a quadratic equation is
with solutions 
.
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
| 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. 
