Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1967 AHSME Problems/Problem 35"

(Created page with "== Problem == The roots of <math>64x^3-144x^2+92x-15=0</math> are in arithmetic progression. The difference between the largest and smallest roots is: <math>\textbf{(A)}\ 2\qqu...")
 
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{B}</math>
+
 
 +
By Vieta, the sum of the roots is <math>-\frac{-144}{64} = \frac{9}{4}</math>.  Because the roots are in arithmetic progression, the middle root is the average of the other two roots, and is also the average of all three roots.  Therefore, <math>\frac{\frac{9}{4}}{3} = \frac{3}{4}</math> is the middle root.
 +
 
 +
The other two roots have a sum of <math>\frac{9}{4} - \frac{3}{4} = \frac{3}{2}</math>.  By Vieta on the original cubic, the product of all <math>3</math> roots is <math>-\frac{-15}{64} = \frac{15}{64}</math>, so the product of the remaining two roots is <math>\frac{\frac{15}{64}}{\frac{3}{4}} = \frac{5}{16}</math>.  If the sum of the two remaining roots is <math>\frac{3}{2} = \frac{24}{16}</math>, and the product is <math>\frac{5}{16}</math>, the roots are also the two roots of <math>16x^2 - 24x + 5 = 0</math>.
 +
 
 +
The two remaining roots are <math>x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}</math>, and they have a difference of <math>\frac{2\sqrt{B^2 - 4AC}{2A}</math>.  Plugging in gives <math>\frac{\sqrt{24^2 - 4(16)(5)}{16}</math>, which is equal to <math>1</math>, which is answer <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:15, 12 July 2019

Problem

The roots of $64x^3-144x^2+92x-15=0$ are in arithmetic progression. The difference between the largest and smallest roots is:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ \frac{1}{2}\qquad \textbf{(D)}\ \frac{3}{8}\qquad \textbf{(E)}\ \frac{1}{4}$

Solution

By Vieta, the sum of the roots is $-\frac{-144}{64} = \frac{9}{4}$. Because the roots are in arithmetic progression, the middle root is the average of the other two roots, and is also the average of all three roots. Therefore, $\frac{\frac{9}{4}}{3} = \frac{3}{4}$ is the middle root.

The other two roots have a sum of $\frac{9}{4} - \frac{3}{4} = \frac{3}{2}$. By Vieta on the original cubic, the product of all $3$ roots is $-\frac{-15}{64} = \frac{15}{64}$, so the product of the remaining two roots is $\frac{\frac{15}{64}}{\frac{3}{4}} = \frac{5}{16}$. If the sum of the two remaining roots is $\frac{3}{2} = \frac{24}{16}$, and the product is $\frac{5}{16}$, the roots are also the two roots of $16x^2 - 24x + 5 = 0$.

The two remaining roots are $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, and they have a difference of $\frac{2\sqrt{B^2 - 4AC}{2A}$ (Error compiling LaTeX. Unknown error_msg). Plugging in gives $\frac{\sqrt{24^2 - 4(16)(5)}{16}$ (Error compiling LaTeX. Unknown error_msg), which is equal to $1$, which is answer $\fbox{B}$.

See also

1967 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1967_AHSME_Problems/Problem_35&oldid=107579"