How to go from solutions to an equation? I need help to find an equation with the following roots: x=1+2i ; x=1+2i ; x= 1-2i ; x=1-2i . P.S. they are repeated so all I know is a 4th grade equation.

Question

jango_in_dtown · Answer

[{x-(1+2i)}{x-(1-2i)}]^2=0
or, [{(x-1)-2i}{(x-1)+2i}]^2=0
or, {(x-1)^2-(2i)^2}^2=0
or, {(x-1)^2+4}^2=0

jango_in_dtown · Answer

or, (x^2-2x+5)^2=0

nvafer · Answer

{(x-1)^2+4}^2=0 how did you get to this one specifically @jango_IN_DTOWN ?

jango_in_dtown · Answer

(a+b)(a-b)=a^2-b^2 and -(2i)^2=-(4i^2)=-4(-1)=4

jango_in_dtown · Answer

a=x-1, b= 2i

sshayer · Answer

$$\left\{ x-\left( 1+2i \right) \right\}^2\left\{ x-\left( 1-2i \right) \right\}^2$$
$$=\left[ \left\{ x-\left( 1+2i \right) \right\}\left\{ x-\left( 1-2i \right) \right\} \right]^2$$
$$=\left[ \left\{ \left( x-1 \right)-2i \right\}\left\{ \left( x-1 \right)+2i \right\} \right]^2$$
$$=\left[ \left( x-1 \right)^2-\left( 2i \right)^2 \right]$$
$$=\left\{ x^2- 2x+1-4i^2 \right\}^2=\left( x^2-2x+1+4 \right)^2=\left( x^2-2x+5 \right)^2$$

nvafer · Answer

Terrific guys @jango_IN_DTOWN  @sshayer  thanks

sshayer · Answer

yw