View source for 1959 AHSME Problems/Problem 44

== Problem ==

The roots of <math>x^2+bx+c=0</math> are both real and greater than <math>1</math>. Let <math>s=b+c+1</math>. Then <math>s</math>:
<math>\textbf{(A)}\ \text{may be less than zero}\qquad\textbf{(B)}\ \text{may be equal to zero}\qquad </math>

<math>\textbf{(C)} \text{ must be greater than zero}\qquad\textbf{(D)}\ \text{must be less than zero}\qquad 
\textbf{(E)}\text{ must be between -1 and +1}</math>   

== Solution ==

Let the roots of the quadratic be <math>p</math> and <math>q</math>. Then, by [[Vieta's Formulas]], <math>b=-(p+q)</math> and <math>c=pq</math>. By substituting these values of <math>b</math> and <math>c</math> into our expression for <math>s</math>, we see that <math>s=pq-p-q+1</math>. By [[SFFT]], <math>s=(p-1)(q-1)</math>. From the problem, we know that <math>p</math> and <math>q</math> are both greater than <math>1</math>, so <math>(p-1)</math> and <math>(q-1)</math> are necessarily positive. Thus, <math>s</math>, the product of these two positive terms, <math>\fbox{\textbf{(C) }must be greater than zero}</math>.

== See also ==
{{AHSME 50p box|year=1959|num-b=43|num-a=45}}
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