If the roots of `6x^3 -11 x^2 +6x-1=0` are in H.P then one of the roots is

Options

A. `1/2`
B. `-1/2`
C. `2`
D. `-2`

Answer

To solve the problem, we need to find one of the roots of the polynomial equation \(6x^3 - 11x^2 + 6x - 1 = 0\) given that its roots are in Harmonic Progression (H.P.). We will follow these steps:

### Step 1: Understanding the Relationship Between H.P. and A.P.
If the roots of the polynomial are in Harmonic Progression, then the reciprocals of the roots will be in Arithmetic Progression (A.P.).

### Step 2: Let the Roots be \(a, b, c\)
Let the roots of the polynomial be \(r_1, r_2, r_3\). Since they are in H.P., we can express their reciprocals as:
\[ \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3} \]
These are in A.P., so we can denote them as:
\[ \frac{1}{r_1} = a - d, \quad \frac{1}{r_2} = a, \quad \frac{1}{r_3} = a + d \]

### Step 3: Sum of the Roots
Using Vieta's formulas, we know that the sum of the roots \(r_1 + r_2 + r_3\) is given by:
\[ r_1 + r_2 + r_3 = \frac{-(-11)}{6} = \frac{11}{6} \]
In terms of \(a\) and \(d\):
\[ \frac{1}{(a - d)} + \frac{1}{a} + \frac{1}{(a + d)} = \frac{11}{6} \]

### Step 4: Product of the Roots
The product of the roots \(r_1 r_2 r_3\) is given by:
\[ r_1 r_2 r_3 = \frac{-(-1)}{6} = \frac{1}{6} \]
In terms of \(a\) and \(d\):
\[ \left(a - d\right) a \left(a + d\right) = \frac{1}{6} \]

### Step 5: Simplifying the Equations
From the sum of the roots:
\[ \frac{(a - d) + a + (a + d)}{(a - d)a(a + d)} = \frac{11}{6} \]
This simplifies to:
\[ \frac{3a}{(a^2 - d^2)} = \frac{11}{6} \]
Cross-multiplying gives:
\[ 18a = 11(a^2 - d^2) \]

From the product of the roots:
\[ a^3 - ad^2 = \frac{1}{6} \]

### Step 6: Solving the System of Equations
Now we have two equations:
1. \(18a = 11a^2 - 11d^2\)
2. \(a^3 - ad^2 = \frac{1}{6}\)

We can express \(d^2\) from the first equation and substitute it into the second equation to find \(a\).

### Step 7: Finding Roots
After solving the equations, we can find the values of \(a\) and \(d\). The roots will be:
\[ r_1 = \frac{1}{(a - d)}, \quad r_2 = \frac{1}{a}, \quad r_3 = \frac{1}{(a + d)} \]

### Conclusion
After solving, we find that one of the roots of the original polynomial \(6x^3 - 11x^2 + 6x - 1 = 0\) is \(\frac{1}{2}\).