If 2,3 are two roots of the reciprocal equation `6x^5 - 29 x^4 +2x^3 +2x^2 -29 x+6=0` then the other roots are

Options

A. a) `1,-2,3`
B. b) `-1,-2,-3`
C. c) `-1,1/2,1/3`
D. d) `1,1/2,1/3`

Answer

To find the other roots of the given reciprocal equation \( 6x^5 - 29x^4 + 2x^3 + 2x^2 - 29x + 6 = 0 \) where \( 2 \) and \( 3 \) are roots, we can follow these steps:

### Step 1: Identify the Reciprocal Roots
Since \( 2 \) and \( 3 \) are roots of the reciprocal equation, the roots of the original equation will be their reciprocals. Therefore, we have:
\[ \text{Reciprocal roots} = \frac{1}{2}, \frac{1}{3} \]

### Step 2: Set Up the Reciprocal Equation
To find the reciprocal equation, we substitute \( x = \frac{1}{y} \) into the original equation:
\[ 6\left(\frac{1}{y}\right)^5 - 29\left(\frac{1}{y}\right)^4 + 2\left(\frac{1}{y}\right)^3 + 2\left(\frac{1}{y}\right)^2 - 29\left(\frac{1}{y}\right) + 6 = 0 \]

### Step 3: Multiply Through by \( y^5 \)
Multiplying the entire equation by \( y^5 \) to eliminate the fractions gives:
\[ 6 - 29y + 2y^2 + 2y^3 - 29y^4 + 6y^5 = 0 \]
Rearranging this, we get:
\[ 6y^5 - 29y^4 + 2y^3 + 2y^2 - 29y + 6 = 0 \]

### Step 4: Finding Other Roots
Since we already have two roots \( \frac{1}{2} \) and \( \frac{1}{3} \), we need to find the other roots. We can test some simple values to find additional roots. Let's test \( y = -1 \):
\[ 6(-1)^5 - 29(-1)^4 + 2(-1)^3 + 2(-1)^2 - 29(-1) + 6 \]
Calculating this:
\[ 6(-1) - 29(1) + 2(-1) + 2(1) + 29 + 6 = -6 - 29 - 2 + 2 + 29 + 6 = 0 \]
Since the equation equals zero, \( y = -1 \) is indeed a root.

### Step 5: Conclusion
Thus, the roots of the original equation are:
\[ \frac{1}{2}, \frac{1}{3}, -1 \]

### Final Answer
The other roots of the equation are \( -1, \frac{1}{2}, \frac{1}{3} \).

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