1967 AHSME Problems/Problem 35

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'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 $\fbox{B}$ .

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 34	Followed by Problem 36
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1967 AHSME Problems/Problem 35

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