# How do you write a possible quadratic equation in standard form that has the given roots {1+or-i}?

Mar 29, 2018

The possible cubic equation happens to be:

${x}^{3} - {x}^{2} + x - 1 = 0$

#### Explanation:

roots are
$x = 1$

$x - 1 = 0$

$x = - i$

$x + i = 0$

Complex numbers occur always in conjugates

$x = i$,
$x - i = 0$

$x = - i$,
$x + i = 0$

Hence, the factors are:

$\left(x - 1\right) , \left(x - i\right) , \left(x + i\right)$

$\left(x - 1\right) \left(x - i\right) \left(x + i\right) = 0$

$\left(x - 1\right) \left({x}^{2} + 1\right) = 0$

The possible cubic equation happens to be:

${x}^{3} - {x}^{2} + x - 1 = 0$