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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}}.</math>
+
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}}

Latest revision as of 20:30, 30 July 2019

Problem 34

Let the roots of $x^2-3x+1=0$ be $r$ and $s$. Then the expression $r^2+s^2$ is: $\textbf{(A)}\ \text{a positive integer} \qquad\textbf{(B)}\ \text{a positive fraction greater than 1}\qquad\textbf{(C)}\ \text{a positive fraction less than 1}\qquad\textbf{(D)}\ \text{an irrational number}\qquad\textbf{(E)}\ \text{an imaginary number}$

Solution

You may recognize that $r^2+s^2$ can be written as $(r+s)^2-2rs$. Then, by Vieta's formulas, \[r+s=-(-3)=3\]and \[rs=1.\]Therefore, plugging in the values for $r+s$ and $rs$, \[(r+s)^2-2rs=(3)^2-2(1)=9-2=7.\]Hence, we can say that the expression $r^2+s^2$ is $\boxed{\text{(A)}\ \text{a positive integer.}}$

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 33 		Followed by
Problem 35
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All AHSME Problems and Solutions

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