Write  a quadratic equation whose root is (1-i)(2+i).

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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Let the first root be z such that:

z= (1-i)(2+i)

  = 2 +i -2i -i^2

  = 3 -i

The the first root is:

z= 3-i

The second root would be z' such that:

z'= 3+i

Now to find the equation:

z+z' = 3-i + 3 +i = 6

z*z' = (3-i)(3+i) = 9 +1 = 10

Then the equation is:

f(x) = x^2 -6x + 10

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neela | High School Teacher | (Level 3) Valedictorian

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The given root is (1-i)(2+i) = 2+i - 2i +1 = 3-i which is an imaginary root.

A quadratic equation , if it has imaginary roots, has them in pairs and each conjugates .

Therefore the quadratic has  the other root 3+i, the conjugate of 3-i.

So the required quadratic equation with roots 3+i and 3-i is

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

(x-3)^2  - i^2

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

x^2-6x+10 = 0

Top Answer

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If an equation has a complex root, then the equation has another root, namely the conjugate of the complex root.

Because it is a quadratic equation, the number of roots is 2, and we have already one root, so the second root is the conjugate of the number (1-i)(2+i).

We'll calculate (1-i)(2+i).

(1-i)(2+i) = 2 + i - 2i - i^2, where i^2 = -1

(1-i)(2+i) = 2 - i + 1

(1-i)(2+i) = 3 - i

The first root is z1 = 3 - i.

The conjugate of the complex number z1 is the second root,

z2 = 3 + i.

We'll form the quadratic equation using Viete's relations:

z1 + z2 = 3 - i + 3 + i

We'll combine like terms:

z1 + z2 = 6

z1*z2 = (3-i)(3+i) = 9 - i^2

 z1*z2 = 9+1

z1*z2 = 10

The quadratic equation is:

z^2 - (z1+z2)*z + (z1*z2) = 0

z^2 - 6z + 10 = 0

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