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

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Answer 1 · 2018-03-29T06:09:08

The possible cubic equation happens to be:

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

roots are
$x = 1$ $x=1$

$x - 1 = 0$ $x-1=0$

$x = - i$ $x=-i$

$x + i = 0$ $x+i=0$

Complex numbers occur always in conjugates

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

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

Hence, the factors are:

$(x - 1), (x - i), (x + i)$ $(x-1), (x-i), (x+i)$

$(x - 1) (x - i) (x + i) = 0$ $(x-1)(x-i)(x+i)=0$

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

The possible cubic equation happens to be:

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