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Difference between revisions of "1959 AHSME Problems/Problem 44"

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== Solution ==
 
== Solution ==
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<math>\fbox{C}</math>
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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 ==
 
== See also ==
 
{{AHSME 50p box|year=1959|num-b=41|num-a=43}}
 
{{AHSME 50p box|year=1959|num-b=41|num-a=43}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:25, 22 July 2024

Problem

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

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

Solution

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

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 41 		Followed by
Problem 43
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All AHSME Problems and Solutions

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