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1967 AMC 12 Problem 35

Attempt Problem 35 of the 1967 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1967 AMC 12 solutions, or check the answer key.

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35.

The roots of 64x3−144x2+92x−15=064x^3-144x^2+92x-15=0 are in arithmetic progression. The difference between the largest and smallest roots is:

22

11

12\dfrac12

38\dfrac38

14\dfrac14

Answer: B
Concepts:polynomialarithmetic sequenceVieta’s Formulas
Difficulty rating: 1750
Small Hint:

Write the roots as t−d,t-d, t,t, and t+dt+d

Big Hint:

Use their sum to find t,t, then use their product to find d2d^2

Solution:

Let the roots be t−d,t-d, t,t, and t+d.t+d. Their sum is 3t=14464=94,3t=\frac{144}{64}=\frac{9}{4}, so t=34.t=\frac{3}{4}. Their product is t(t2−d2)=1564. t(t^2-d^2)=\frac{15}{64}. Substituting t=34t=\frac{3}{4} gives d2=14,d^2=\frac{1}{4}, so the difference between the extreme roots is 2∣d∣=1.2\lvert d\rvert=1.

Therefore, the correct answer is B.

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