Multiple choice

Solve the equation $\displaystyle 6x^{3}-11x^{2}+6x-1= 0$ and find the smallest of roots.Given the roots are in harmonical progression.

  1. 1

  2. 1/3

  3. 1/2

  4. 1/4
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B Correct answer
Explanation

If roots are in harmonic progression, their reciprocals are in arithmetic progression. Let roots be 1/a, 1/b, 1/c. The equation 6x^3 - 11x^2 + 6x - 1 = 0 has roots 1/1, 1/2, 1/3. The smallest root is 1/3.

AI explanation

If the roots of the cubic equation 6x cubed - 11x squared + 6x - 1 = 0 are in harmonic progression, their reciprocals must be in arithmetic progression. Let the roots be 1/a, 1/(a+d), and 1/(a+2d); substituting y = 1/x transforms the equation into y cubed - 6y squared + 11y - 6 = 0. Since the y values form an arithmetic sequence, their sum is 6, making the middle term 6 divided by 3, which is 2. The sequence is 1, 2, 3, meaning the original x roots are 1, 1/2, and 1/3. The smallest of these roots is 1/3.

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