Solve the following quadratic equation: `(2+i)x^2-(5-i)x+2(1-i)=0`

Explanation

To solve the quadratic equation \( (2+i)x^2 - (5-i)x + 2(1-i) = 0 \), we will follow these steps:

### Step 1: Identify coefficients
We can compare the given quadratic equation with the standard form \( ax^2 + bx + c = 0 \) to identify the coefficients \( a \), \( b \), and \( c \).

- \( a = 2 + i \)
- \( b = -(5 - i) = -5 + i \)
- \( c = 2(1 - i) = 2 - 2i \)

### Step 2: Calculate the discriminant
The discriminant \( D \) is given by the formula:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-5 + i)^2 - 4(2 + i)(2 - 2i) \]

Calculating \( (-5 + i)^2 \):
\[ (-5 + i)^2 = 25 - 10i - 1 = 24 - 10i \]

Calculating \( 4(2 + i)(2 - 2i) \):
\[ (2 + i)(2 - 2i) = 4 - 4i + 2i - 2i^2 = 4 - 2i + 2 = 6 - 2i \]
Thus,
\[ 4(2 + i)(2 - 2i) = 4(6 - 2i) = 24 - 8i \]

Now substituting back into the discriminant:
\[ D = (24 - 10i) - (24 - 8i) = -10i + 8i = -2i \]

### Step 3: Find the roots using the quadratic formula
The roots of the quadratic equation are given by:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ x = \frac{-(-5 + i) \pm \sqrt{-2i}}{2(2 + i)} = \frac{5 - i \pm \sqrt{-2i}}{4 + 2i} \]

### Step 4: Simplify \( \sqrt{-2i} \)
To find \( \sqrt{-2i} \), we can express it in the form \( a + bi \):
Let \( \sqrt{-2i} = a + bi \). Then,
\[ (-2i) = (a + bi)^2 = a^2 - b^2 + 2abi \]
This gives us the equations:
1. \( a^2 - b^2 = 0 \) (real part)
2. \( 2ab = -2 \) (imaginary part)

From the first equation, \( a^2 = b^2 \) implies \( a = \pm b \). Substituting \( b = a \) into the second equation:
\[ 2a^2 = -2 \implies a^2 = -1 \implies a = \pm i \]
Thus, \( \sqrt{-2i} = \pm (1 - i) \).

### Step 5: Substitute back to find the roots
Using \( \sqrt{-2i} = 1 - i \):
\[ x = \frac{5 - i \pm (1 - i)}{4 + 2i} \]
Calculating the two cases:
1. \( x_1 = \frac{5 - i + 1 - i}{4 + 2i} = \frac{6 - 2i}{4 + 2i} \)
2. \( x_2 = \frac{5 - i - (1 - i)}{4 + 2i} = \frac{4}{4 + 2i} \)

### Step 6: Simplify the fractions
For \( x_1 \):
\[ x_1 = \frac{6 - 2i}{4 + 2i} \cdot \frac{4 - 2i}{4 - 2i} = \frac{(6 - 2i)(4 - 2i)}{16 + 4} = \frac{24 - 12i - 8i + 4}{20} = \frac{28 - 20i}{20} = 1.4 - i \]

For \( x_2 \):
\[ x_2 = \frac{4}{4 + 2i} \cdot \frac{4 - 2i}{4 - 2i} = \frac{16 - 8i}{16 + 4} = \frac{16 - 8i}{20} = 0.8 - 0.4i \]

### Final Roots
Thus, the roots of the quadratic equation are:
\[ x_1 = 1.4 - i \quad \text{and} \quad x_2 = 0.8 - 0.4i \]