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

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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, 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 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 <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>.
  
 
== See also ==
 
== See also ==

Latest revision as of 14:17, 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 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 $\fbox{B}$.

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 34
Followed by
Problem 36
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