Write a quadratic equation that has 5 + 2i as one of its roots.

Can someone show me a step by step method on solving the above problem.

2. Originally Posted by cityismine
Write a quadratic equation that has 5 + 2i as one of its roots.

Can someone show me a step by step method on solving the above problem.

0 = (x - [5 + 2i])(ax + b) where a and b are any numbers you please ......

However, if you want your equation to have real coefficients, then

$0 = (x - [5 + 2i])(x - [5 - 2i]) = ([x - 5] - 2i)([x - 5] + 2i) = [x - 5]^2 + 4 = ........$

3. Mr. fantastic can you explain the second method of "real coefficients" a bit more thoroughly. I'm still a little confused how you got the initial equation.

4. Originally Posted by cityismine
Mr. fantastic can you explain the second method of "real coefficients" a bit more thoroughly. I'm still a little confused how you got the initial equation.
The only way to make disappear such a complex number is to multiply by its conjugate.

Let $\bar{z}$ be the conjugate of z.

$(x-z)(x-\bar{z})=x^2-x(\underbrace{z+\bar{z}}_{\text{real number}})+\underbrace{z \cdot \bar{z}}_{\text{real number}}$

See why these are real numbers by noting that $z=a+ib$ (therefore, $\bar{z}=a-ib$)