# Complex numbers equation

How to solve $x^4-(5-2i)x^2-10i=0$ with complex numbers?

 galactus Posts: 2,272, Reputation: 1436 Ultra Member #2 Nov 13, 2010, 10:04 AM
We can factor just like we would a regular old quadratic.

$x^{4}-(5-2i)x^{2}-10i=0$

What two numbers when multiplied equal -10i and when added equal -5+2i?.

$x^{4}+2ix^{2}-5x^{2}-10i$

Group and factor

$x^{2}(x^{2}+2i)-5(x^{2}+2i)$

$(x^{2}-5)(x^{2}+2i)=0$

What values of x make

$x^{2}-5=0$?.

Difference of two squares:

$(x+\sqrt{5})(x-\sqrt{5})=0$

What values of x make

$x^{2}+2i=0$?.

Factor:

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

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