1959 AHSME Problems/Problem 34

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 $\boxed{\text{(A)}\ \text{a positive integer}}.$

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 33	Followed by Problem 35
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1959 AHSME Problems/Problem 34

Problem 34

Solution

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