Difference between revisions of "1967 AHSME Problems/Problem 18"

(Created page with "== Problem == If <math>x^2-5x+6<0</math> and <math>P=x^2+5x+6</math> then <math>\textbf{(A)}\ P \; \text{can take any real value} \qquad \textbf{(B)}\ 20<P<30\\ \textbf{(C)}\ 0<...")
 
 
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== Solution ==
 
== Solution ==
<math>\fbox{B}</math>
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We are given that <math>x^2 - 5x + 6 < 0</math>, which, when factored, gives <math>(x - 2)(x-3) < 0</math>.  This has a solution of <math>2<x<3</math>, because the original quadratic is <math>\cup</math>-shaped, and thus dips below the x-axis between the roots.
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Since <math>x^2 + 5x + 6</math> has a vertex minimum at <math>x = -\frac{5}{2}</math>, so it is increasing on the interval <math>[2, 3]</math>.  Thus, evaluating <math>P</math> at <math>x=2</math> and <math>x=3</math> will give our bounds, and doing so gives <math>20 < P < 30</math>, or <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 02:09, 13 July 2019

Problem

If $x^2-5x+6<0$ and $P=x^2+5x+6$ then

$\textbf{(A)}\ P \; \text{can take any real value} \qquad \textbf{(B)}\ 20<P<30\\ \textbf{(C)}\ 0<P<20 \qquad \textbf{(D)}\ P<0 \qquad \textbf{(E)}\ P>30$

Solution

We are given that $x^2 - 5x + 6 < 0$, which, when factored, gives $(x - 2)(x-3) < 0$. This has a solution of $2<x<3$, because the original quadratic is $\cup$-shaped, and thus dips below the x-axis between the roots.

Since $x^2 + 5x + 6$ has a vertex minimum at $x = -\frac{5}{2}$, so it is increasing on the interval $[2, 3]$. Thus, evaluating $P$ at $x=2$ and $x=3$ will give our bounds, and doing so gives $20 < P < 30$, or $\fbox{B}$.

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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