Difference between revisions of "1959 AHSME Problems/Problem 34"
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Let the roots of <math>x^2-3x+1=0</math> be <math>r</math> and <math>s</math>. Then the expression <math>r^2+s^2</math> is: <math>\textbf{(A)}\ \text{a positive integer} \qquad\textbf{(B)}\ \text{a positive fraction greater than 1}\qquad\textbf{(C)}\ \text{a positive fraction less than 1}\qquad\textbf{(D)}\ \text{an irrational number}\qquad\textbf{(E)}\ \text{an imaginary number}</math> | Let the roots of <math>x^2-3x+1=0</math> be <math>r</math> and <math>s</math>. Then the expression <math>r^2+s^2</math> is: <math>\textbf{(A)}\ \text{a positive integer} \qquad\textbf{(B)}\ \text{a positive fraction greater than 1}\qquad\textbf{(C)}\ \text{a positive fraction less than 1}\qquad\textbf{(D)}\ \text{an irrational number}\qquad\textbf{(E)}\ \text{an imaginary number}</math> | ||
==Solution== | ==Solution== | ||
| − | You may recognize that <math>r^2+s^2</math> can be written as <math>(r+s)^2-2rs</math>. Then, by [[Vieta's formulas]], <cmath>r+s=-(-3)=3</cmath>and <cmath>rs=1.</cmath>Therefore, plugging in the values for <math>r+s</math> and <math>rs</math>, <cmath>(r+s)^2-2rs=(3)^2-2(1)=9-2=7.</cmath>Hence, we can say that the expression <math>r^2+s^2</math> is <math>\boxed{\text{(A)}\ \text{a positive integer}} | + | You may recognize that <math>r^2+s^2</math> can be written as <math>(r+s)^2-2rs</math>. Then, by [[Vieta's formulas]], <cmath>r+s=-(-3)=3</cmath>and <cmath>rs=1.</cmath>Therefore, plugging in the values for <math>r+s</math> and <math>rs</math>, <cmath>(r+s)^2-2rs=(3)^2-2(1)=9-2=7.</cmath>Hence, we can say that the expression <math>r^2+s^2</math> is <math>\boxed{\text{(A)}\ \text{a positive integer.}}</math> |
==See also== | ==See also== | ||
{{AHSME 50p box|year=1959|num-b=33|num-a=35}} | {{AHSME 50p box|year=1959|num-b=33|num-a=35}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:30, 30 July 2019
Problem 34
Let the roots of 
be 
and 
. Then the expression 
is: 
Solution
You may recognize that can be written as 
. Then, by Vieta's formulas, ![\[r+s=-(-3)=3\]](http://latex.artofproblemsolving.com/9/0/c/90caadf9fc7dc8f9380723add0e1ba1ebf4f943e.png)
and ![\[rs=1.\]](http://latex.artofproblemsolving.com/1/f/a/1faeb986344396c94b1620cd7ea4a59ebc6774dc.png)
Therefore, plugging in the values for 
and 
, ![\[(r+s)^2-2rs=(3)^2-2(1)=9-2=7.\]](http://latex.artofproblemsolving.com/7/3/8/738e147e24888d0718f2afe5741ecd7f42a58179.png)
Hence, we can say that the expression is 
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 33 |
Followed by Problem 35 | |
| 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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
