# quadratic equation.Find the quadratic equation that has one root (1-i)(2+i) .

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The quadratic equation has one root (1-i)(2+i) or (2 + i - 2i - i^2) or 3 - i

Complex roots in a quadratic equation are always conjugates. So the other root is 3 + i

The equation is (x - (3 - i))(x - (3 + i)) = 0

=> (x - 3 + i))(x - 3 - i)) = 0

=> (x - 3)^2 - i^2 = 0

=> x^2 + 9 - 6x + 1 = 0

=> x^2 - 6x + 10 = 0

**The quadratic equation with a root (1-i)(2+i) is x^2 - 6x + 10 = 0**

The quadratic equation has the following form:

ax^2 + bx + c = 0

We'll put x1 = (1-i)(2+i)

We'll remove the brackets:

x1 = 2 + i - 2i - i^2

We'll combine imaginary terms:

x1 = 2 - i - i^2

We'll substitute i^2 = -1

x1 = 2 - i + 1

x1 = 3 - i

The other root is the conjugate of x1:

x2 = 3 + i

The quadratic equation is:

(x - x1)(x - x2) = 0

We'll substitute x1 and x2:

(x - 3 + i)(x - 3 - i) = 0

(x - 3)^2 - i^2 = 0

We'll expand the square:

x^2 - 6x + 9 + 1 = 0

x^2 - 6x + 10 = 0

**The quadratic equation whose roots are 3 - i and 3 + i, is:**

**x^2 - 6x + 10 = 0**