If the roots of ` 24x^3 - 26 x^2 + 9x -1=0` are in H.P then the roots are

Options

A. `1/2,1/3,1/4`
B. `1,1/3,1/8`
C. `1/8,1/3,1/3`
D. `2,3,4`

Answer

To solve the equation \( 24x^3 - 26x^2 + 9x - 1 = 0 \) 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 are in H.P., then their reciprocals will be in Arithmetic Progression (A.P.). Let the roots be \( a - d, a, a + d \).

### Step 2: Substitute \( x = \frac{1}{y} \)
To find the new polynomial in terms of \( y \), we substitute \( x = \frac{1}{y} \):
\[ 24\left(\frac{1}{y}\right)^3 - 26\left(\frac{1}{y}\right)^2 + 9\left(\frac{1}{y}\right) - 1 = 0 \]
This simplifies to:
\[ \frac{24}{y^3} - \frac{26}{y^2} + \frac{9}{y} - 1 = 0 \]
Multiplying through by \( y^3 \) to eliminate the denominators gives:
\[ 24 - 26y + 9y^2 - y^3 = 0 \]
Rearranging this, we have:
\[ -y^3 + 9y^2 - 26y + 24 = 0 \]
or equivalently:
\[ y^3 - 9y^2 + 26y - 24 = 0 \]

### Step 3: Coefficients comparison
Now we compare this with the general cubic equation \( ay^3 + by^2 + cy + d = 0 \):
- \( a = 1 \)
- \( b = -9 \)
- \( c = 26 \)
- \( d = -24 \)

### Step 4: Finding the sum of the roots
Using Vieta's formulas, the sum of the roots \( (a - d) + a + (a + d) = 3a \) is equal to \( -\frac{b}{a} = 9 \):
\[ 3a = 9 \implies a = 3 \]

### Step 5: Finding the product of the roots
The product of the roots \( (a - d) \cdot a \cdot (a + d) = a^3 - ad^2 \) is equal to \( -\frac{d}{a} = 24 \):
\[ 3^3 - 3d^2 = 24 \implies 27 - 3d^2 = 24 \implies 3d^2 = 3 \implies d^2 = 1 \implies d = \pm 1 \]

### Step 6: Finding the roots
Now we can find the roots:
- \( a - d = 3 - 1 = 2 \)
- \( a = 3 \)
- \( a + d = 3 + 1 = 4 \)

### Step 7: Roots in H.P.
The roots in H.P. are the reciprocals of the roots in A.P.:
\[ \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \]

### Final Answer
Thus, the roots of the equation \( 24x^3 - 26x^2 + 9x - 1 = 0 \) are:
\[ \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \]