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

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Given one root of the quadratic equation is (1-i)(2+i) .

(1-i)(2+i) = 2+i-2i-i^2 = 2-i+1 , as i^2 = -1.

=> (1-i)(2+i) = 3 - i is the given root.

Therefore the other root is 3+i, asÂ complex rootsÂ of a polynomial is always in pairs of conjugates.

So if z1 and z2 are the roots of a quadratic, then the quadratic equation is (x-z1)(x-z2) = 0.

=> (x- (3-i))(x-(3+i)) = 0 is the required equation.

=> {(x-3)+i}{x-3)-i} = 0.

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

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

= > x^2-6x+10 = 0 is the required quadratic equation.

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**