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"

m (→‎See also)
(→‎Solution 2)
 
(3 intermediate revisions by 2 users not shown)
Line 8: Line 8:
 
\textbf{(E)}\ \frac{1}{4}</math>
 
\textbf{(E)}\ \frac{1}{4}</math>
  
−
== Solution ==
+
== Solution 1 ==
  
−
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.
+
By [[Vieta's Formulas]], 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 two remaining roots are also the two roots of <math>16x^2 - 24x + 5 = 0</math>.
 
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 two remaining roots are also the two roots of <math>16x^2 - 24x + 5 = 0</math>.
  
 
The two remaining roots are thus <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>.
 
The two remaining roots are thus <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>.
 +
 +
== Solution 2 ==
 +
 +
Because the roots are in arithmetic progression, let the roots be <math>r-d, r,</math> and <math>r+d</math> with <math>d \geq 0</math>. From [[Vieta's Formulas]], we know that <math>(r-d)+r+(r+d)=\tfrac{144}{64}</math>, so <math>3r=\tfrac 9 4</math>, and <math>r=\tfrac 3 4</math>.
 +
 +
By using Vieta again, we see that <math>(r-d)(r)(r+d)=\tfrac{15}{64}</math>. Substituting <math>r=\tfrac 3 4</math> into this equation, we can now solve for <math>d</math>:
 +
\begin{align*}
 +
(\frac 3 4-d)(\frac 3 4)(\frac 3 4+d) &= \frac{15}{64} \\
 +
\frac 3 4(\frac 9{16}-d^2) &= \frac{15}{64} \\
 +
\frac 9{16}-d^2 &= \frac 5{16} \\
 +
d^2 &= \frac 4{16} = \frac 1 4 \\
 +
d &= \pm \frac 1 2
 +
\end{align*}
 +
Because we defined <math>d \geq 0</math>, <math>d=\frac 1 2</math>. Therefore, the difference between the largest and smallest roots, <math>r+d</math> and <math>r-d</math>, is <math>2d=2 \cdot \tfrac 1 2 = \boxed{\textbf{(B) }1}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 19:45, 26 August 2024

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 1

By Vieta's Formulas, 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 two remaining roots are also the two roots of $16x^2 - 24x + 5 = 0$.

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

Solution 2

Because the roots are in arithmetic progression, let the roots be $r-d, r,$ and $r+d$ with $d \geq 0$. From Vieta's Formulas, we know that $(r-d)+r+(r+d)=\tfrac{144}{64}$, so $3r=\tfrac 9 4$, and $r=\tfrac 3 4$.

By using Vieta again, we see that $(r-d)(r)(r+d)=\tfrac{15}{64}$. Substituting $r=\tfrac 3 4$ into this equation, we can now solve for $d$: \begin{align*} (\frac 3 4-d)(\frac 3 4)(\frac 3 4+d) &= \frac{15}{64} \\ \frac 3 4(\frac 9{16}-d^2) &= \frac{15}{64} \\ \frac 9{16}-d^2 &= \frac 5{16} \\ d^2 &= \frac 4{16} = \frac 1 4 \\ d &= \pm \frac 1 2 \end{align*} Because we defined $d \geq 0$, $d=\frac 1 2$. Therefore, the difference between the largest and smallest roots, $r+d$ and $r-d$, is $2d=2 \cdot \tfrac 1 2 = \boxed{\textbf{(B) }1}$.

See also

1967 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
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=225828"