Question

Question #1527e

Answer 1

Given

`x^2-6x+13 =0`

Substitute `3+2i` for every x:

`(3+2i)^2-6(3+2i)+13 =0`

Perform the multiplication:

`9+12i+4i^2-18-12i+13=0`

Substitute -1 for `i^2`

`9+12i-4-18-12i+13=0`

Combine like terms:

`0 = 0`

This shows that `3+2i` is a root.

The conjugate of `3+2i` is `3-2i`

Substitute `3-2i` for every x:

`(3-2i)^2-6(3-2i)+13 =0`

Perform the multiplication:

`9-12i+4i^2-18+12i+13=0`

Substitute -1 for `i^2`

`9-12i-4-18+12i+13=0`

Combine like terms:

`0 = 0`

This shows that `3-2i` is a root.

Answer 2

See below.

Explanation:

Given a polynomial with real coefficients like

`x^2+ax+b=0` with `{a,b} in RR` we have that it's two roots obey

`(x-x_1)(x-x_2) = x^2-(x_1+x_2)x+x_1x_2=0` or

`{(x_1+x_2 = -a),(x_1 x_2 = b):}`

now supposing

`x_1 = u_1+iv_1`,
`x_2 = u_2+i v_2`

`{(u_1+u_2 +i(v_1+v_2)= -a),((u_1+iv_1)( u_2+iv_2) = b):}`

From the first equation, if `a in RR rArr {(u_1+u_2 = -a),(v_1+v_2 = 0):}`

From the second equation if `b in RR rArr {(u_1u_2-v_1v_2 = b),(u_1v_2+u_2v_1=0):}`

by solving those equations we conclude

`u_1 = u_2` and `v_1 = -v_2` or

`x_1 = u + iv`
`x_2 = u-iv`

so `x_2, x_2` are complex conjugate.