The difference between two roots of the equation `x^3 -13 x^2 + 15 x+189 =0` is 2 then the roots of the equation are

Options

A. `-3,7,9`
B. `-3,-1,-9`
C. `3,5,7`
D. `-3,-1,9`

Answer

To find the roots of the equation \(x^3 - 13x^2 + 15x + 189 = 0\) given that the difference between two of the roots is 2, we can follow these steps:

### Step 1: Define the Roots
Let the roots of the equation be \( \alpha \), \( \alpha + 2 \) (since the difference is 2), and \( \beta \).

### Step 2: Use the Sum of Roots
From Vieta's formulas, the sum of the roots of the polynomial \(x^3 + bx^2 + cx + d = 0\) is given by:
\[ \alpha + (\alpha + 2) + \beta = -\frac{-13}{1} = 13 \]
This simplifies to:
\[ 2\alpha + \beta + 2 = 13 \]
Thus,
\[ 2\alpha + \beta = 11 \quad \text{(Equation 1)} \]

### Step 3: Use the Sum of the Product of Roots Taken Two at a Time
The sum of the products of the roots taken two at a time is given by:
\[ \alpha(\alpha + 2) + \alpha\beta + (\alpha + 2)\beta = 15 \]
Expanding this gives:
\[ \alpha^2 + 2\alpha + \alpha\beta + \alpha\beta + 2\beta = 15 \]
This simplifies to:
\[ \alpha^2 + 2\alpha + 2\alpha\beta + 2\beta = 15 \]

### Step 4: Substitute \(\beta\) from Equation 1
From Equation 1, we have:
\[ \beta = 11 - 2\alpha \]
Substituting this into the equation gives:
\[ \alpha^2 + 2\alpha + 2\alpha(11 - 2\alpha) + 2(11 - 2\alpha) = 15 \]
Expanding this:
\[ \alpha^2 + 2\alpha + 22\alpha - 4\alpha^2 + 22 - 4\alpha = 15 \]
Combining like terms:
\[ -3\alpha^2 + 20\alpha + 22 - 15 = 0 \]
This simplifies to:
\[ -3\alpha^2 + 20\alpha + 7 = 0 \]
Multiplying through by -1 gives:
\[ 3\alpha^2 - 20\alpha - 7 = 0 \quad \text{(Equation 2)} \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \alpha = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 3 \cdot (-7)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ \alpha = \frac{20 \pm \sqrt{400 + 84}}{6} = \frac{20 \pm \sqrt{484}}{6} \]
Since \( \sqrt{484} = 22 \):
\[ \alpha = \frac{20 \pm 22}{6} \]
This gives us two potential values for \( \alpha \):
1. \( \alpha = \frac{42}{6} = 7 \)
2. \( \alpha = \frac{-2}{6} = -\frac{1}{3} \)

### Step 6: Find \(\beta\) for Each \(\alpha\)
Using \( \beta = 11 - 2\alpha \):
1. If \( \alpha = 7 \):
\[ \beta = 11 - 2 \cdot 7 = -3 \]
The roots are \( 7 \), \( 9 \) (since \( \alpha + 2 = 9 \)), and \( -3 \).

2. If \( \alpha = -\frac{1}{3} \):
\[ \beta = 11 - 2 \cdot \left(-\frac{1}{3}\right) = 11 + \frac{2}{3} = \frac{33 + 2}{3} = \frac{35}{3} \]
The roots would not satisfy the original polynomial since they would yield a positive product.

### Conclusion
Thus, the roots of the equation \(x^3 - 13x^2 + 15x + 189 = 0\) are:
\[ \boxed{-3, 7, 9} \]