Difference between revisions of "1969 AHSME Problems/Problem 12"
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== Solution == | == Solution == | ||
| − | <math>\ | + | The quadratic can also be written as |
| + | <cmath>x^2 + \frac{8}{3}x + \frac{m}{2}</cmath> | ||
| + | In order for the quadratic to be a square of a linear expression, the constant term must be the square of the x-term divided by 4. Thus, | ||
| + | <cmath>\frac{m}{2} = (\frac{8}{3})^2 \cdot \frac{1}{4}</cmath> | ||
| + | <cmath>\frac{m}{2} = \frac{16}{9}</cmath> | ||
| + | <cmath>m = \frac{32}{9}</cmath> | ||
| + | The answer is <math>\boxed{\textbf{(A)}}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 03:20, 7 June 2018
Problem
Let 
be the square of an expression which is linear in 
. Then 
has a particular value between:
Solution
The quadratic can also be written as
![\[x^2 + \frac{8}{3}x + \frac{m}{2}\]](http://latex.artofproblemsolving.com/9/1/c/91ca7a85cdbd39b9984abb043324621c8a63b526.png)
In order for the quadratic to be a square of a linear expression, the constant term must be the square of the x-term divided by 4. Thus,
![\[\frac{m}{2} = (\frac{8}{3})^2 \cdot \frac{1}{4}\]](http://latex.artofproblemsolving.com/6/2/f/62fdb73516ad405fffd7358f00ee4b448488fa54.png)
![\[\frac{m}{2} = \frac{16}{9}\]](http://latex.artofproblemsolving.com/f/4/d/f4d7782bc0e4219f7701e72f4e4543f1872d5912.png)
![\[m = \frac{32}{9}\]](http://latex.artofproblemsolving.com/4/3/b/43b244317b632d3badb108b30cb4e7c2bc0d05f0.png)
The answer is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
