Questions: x^3 - 2x^2 + 3x - 6 = 0

Transcript text: 3) $x^{3}-2 x^{2}+3 x-6=0$

Solution

Solution Steps

To solve the cubic equation $x^3 - 2x^2 + 3x - 6 = 0$, we can use numerical methods such as the Newton-Raphson method or a library function to find the roots. Since it's a cubic equation, it may have up to three real roots. We will use Python's numpy library to find the roots of the polynomial.

Step 1: Identify the Roots

The cubic equation $x^3 - 2x^2 + 3x - 6 = 0$ has been analyzed, and the roots have been determined to be: \[ \begin{align_} x_1 & = 2.0000 \\ x_2 & = -9.6945 \times 10^{-16} + 1.7321i \\ x_3 & = -9.6945 \times 10^{-16} - 1.7321i \end{align_} \]

Step 2: Classify the Roots

From the results, we can see that:

$x_1 = 2$ is a real root.
$x_2$ and $x_3$ are complex conjugates, indicating that the cubic equation has one real root and two complex roots.

Final Answer

The roots of the equation $x^3 - 2x^2 + 3x - 6 = 0$ are: \[ \boxed{x = 2} \quad \text{(real root)} \] \[ \boxed{x = -9.6945 \times 10^{-16} + 1.7321i} \quad \text{(complex root)} \] \[ \boxed{x = -9.6945 \times 10^{-16} - 1.7321i} \quad \text{(complex root)} \]

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