For the equilibrium `x^3 +3x^2 -x-2=0` if `s_1 ,s_2 ,s_3` have their usual notaion then

Options

A. a) `s_1 lt _3 lt s_2`
B. b) `s_1 lt s_2 lt s_3`
C. c) `s_2 lt s_1 lt s_3`
D. d) `s_3 lt s_1 lt s_2`

Answer

To solve the equation \( x^3 + 3x^2 - x - 2 = 0 \) and find the values of \( S_1, S_2, S_3 \) (the sums of the roots taken one at a time, two at a time, and three at a time respectively), we can use Vieta's formulas.

### Step-by-Step Solution:

1. **Identify the coefficients of the polynomial:**
The given polynomial is \( x^3 + 3x^2 - x - 2 = 0 \).
Here, the coefficients are:
- \( a = 1 \) (coefficient of \( x^3 \))
- \( b = 3 \) (coefficient of \( x^2 \))
- \( c = -1 \) (coefficient of \( x \))
- \( d = -2 \) (constant term)

2. **Calculate \( S_1 \):**
According to Vieta's formulas, the sum of the roots \( S_1 \) is given by:
\[ S_1 = -\frac{b}{a} = -\frac{3}{1} = -3 \]

3. **Calculate \( S_2 \):**
The sum of the products of the roots taken two at a time \( S_2 \) is given by:
\[ S_2 = \frac{c}{a} = \frac{-1}{1} = -1 \]

4. **Calculate \( S_3 \):**
The product of the roots \( S_3 \) is given by:
\[ S_3 = -\frac{d}{a} = -\frac{-2}{1} = 2 \]

5. **Summarize the results:**
We have found the following values:
- \( S_1 = -3 \)
- \( S_2 = -1 \)
- \( S_3 = 2 \)

6. **Order the sums:**
Now, we can order these sums:
\[ S_1 < S_2 < S_3 \quad \text{(i.e., } -3 < -1 < 2\text{)} \]

7. **Select the correct option:**
Based on the ordering of \( S_1, S_2, S_3 \), we can conclude that:
- \( S_1 \) is the smallest, followed by \( S_2 \), and \( S_3 \) is the largest.

### Conclusion:
The correct ordering is \( S_1 < S_2 < S_3 \) which corresponds to the option that states this relationship.