Difference between revisions of "1959 AHSME Problems/Problem 8"

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<cmath>=4.</cmath>
 
<cmath>=4.</cmath>
 
So, the answer is <math>\boxed{\textbf{(A)} \ 4}</math>.
 
So, the answer is <math>\boxed{\textbf{(A)} \ 4}</math>.
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==See also==
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{{AHSME 50p box|year=1959|num-b=7|num-a=9}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Revision as of 12:03, 21 July 2024

Problem

The value of $x^2-6x+13$ can never be less than:

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 13$

Solution

The $x$ value at which the minimum value of this quadratic occurs is $-\frac{-6}{2\cdot1}=3$. The minimum value of the quadratic is therefore at \[3^2-6\cdot3+13\] \[=9-18+13\] \[=4.\] So, the answer is $\boxed{\textbf{(A)} \ 4}$.

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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

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