Solve `9x^3-15x^2+7x-1=0` , given that two of its roots are equal .

Answer

To solve the cubic equation \(9x^3 - 15x^2 + 7x - 1 = 0\) given that two of its roots are equal, we can follow these steps:

### Step 1: Identify the roots
Given that two roots are equal, we can denote the roots as \( \alpha, \alpha, \beta \).

### Step 2: Use the Rational Root Theorem
We will test possible rational roots. The possible rational roots can be factors of the constant term divided by factors of the leading coefficient. The constant term is \(-1\) and the leading coefficient is \(9\). Therefore, the possible rational roots are \( \pm 1 \).

### Step 3: Test \(x = 1\)
Let's test \(x = 1\):

\[ f(1) = 9(1)^3 - 15(1)^2 + 7(1) - 1 = 9 - 15 + 7 - 1 = 0 \]

Since \(f(1) = 0\), \(x = 1\) is a root.

### Step 4: Factor the polynomial
Since \(x = 1\) is a root, we can factor the polynomial using synthetic division or polynomial long division. We divide \(9x^3 - 15x^2 + 7x - 1\) by \(x - 1\).

Performing synthetic division:

\[ \begin{array}{r|rrrr} 1 & 9 & -15 & 7 & -1 \\ & & 9 & -6 & 1 \\ \hline & 9 & -6 & 1 & 0 \\ \end{array} \]

The result is \(9x^2 - 6x + 1\). Thus, we can write:

\[ 9x^3 - 15x^2 + 7x - 1 = (x - 1)(9x^2 - 6x + 1) \]

### Step 5: Set the quadratic to zero
Now we need to solve \(9x^2 - 6x + 1 = 0\). Since we know that two roots are equal, we can use the condition for equal roots, which is the discriminant must be zero.

### Step 6: Calculate the discriminant
The discriminant \(D\) of the quadratic \(ax^2 + bx + c\) is given by:

\[ D = b^2 - 4ac \]

For \(9x^2 - 6x + 1\):

\[ D = (-6)^2 - 4 \cdot 9 \cdot 1 = 36 - 36 = 0 \]

Since the discriminant is zero, there is one repeated root.

### Step 7: Find the repeated root
Using the formula for the roots of a quadratic equation \(x = \frac{-b \pm \sqrt{D}}{2a}\):

\[ x = \frac{-(-6) \pm 0}{2 \cdot 9} = \frac{6}{18} = \frac{1}{3} \]

### Step 8: List the roots
Thus, the roots of the equation are:

\[ \alpha = \frac{1}{3}, \quad \alpha = \frac{1}{3}, \quad \beta = 1 \]

### Final Answer
The roots of the equation \(9x^3 - 15x^2 + 7x - 1 = 0\) are:

\[ \frac{1}{3}, \frac{1}{3}, 1 \]

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