1959 AHSME Problems/Problem 44

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 43
Followed by
Problem 45
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png