# How do you find the quadratic equation with roots at (2-i) and its complex conjugate?

Jan 9, 2016

Multiply out $\left(x - \left(2 - i\right)\right) \left(x - \left(2 + i\right)\right)$ and simplify to find equation:

${x}^{2} - 4 x + 5 = 0$

#### Explanation:

The Complex conjugate of $2 - i$ is $2 + i$. You get this by inverting the sign of the imaginary part.

Multiply out $\left(x - \left(2 - i\right)\right) \left(x - \left(2 + i\right)\right)$ using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x - 2\right)$ and $b = i$ as follows:

$\left(x - \left(2 - i\right)\right) \left(x - \left(2 + i\right)\right)$

$= \left(\left(x - 2\right) - i\right) \left(\left(x - 2\right) + i\right)$

$= {\left(x - 2\right)}^{2} - {i}^{2}$

$= {x}^{2} - 4 x + 4 + 1$

$= {x}^{2} - 4 x + 5$