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1959 AHSME Problems/Problem 34

Revision as of 07:37, 14 June 2019 by Toinfinity (talk | contribs) (Created page with "==Problem 34== 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...")
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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 $\boxed{\text{(A)}\ \text{a positive integer}}$.

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