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

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== Problem ==
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The value of <math>x^2-6x+13</math> can never be less than:
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<math>\textbf{(A)}\ 4 \qquad
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\textbf{(B)}\ 4.5 \qquad
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\textbf{(C)}\ 5\qquad
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\textbf{(D)}\ 7\qquad
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\textbf{(E)}\ 13  </math> 
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== Solution ==
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The <math>x</math> value at which the minimum value of this quadratic occurs is <math>-\frac{-6}{2\cdot1}=3</math>. The minimum value of the quadratic is therefore at
 
The <math>x</math> value at which the minimum value of this quadratic occurs is <math>-\frac{-6}{2\cdot1}=3</math>. The minimum value of the quadratic is therefore at
 
<cmath>3^2-6\cdot3+13</cmath>
 
<cmath>3^2-6\cdot3+13</cmath>

Revision as of 13:57, 16 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 $\boxed{\textbf{(A)} \ 4}$.

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