Hi, can anyone show me how to do this problem? 100 points for this. in advance

z^2 + (-1 + 5·i)·z + 14 - 7·i = 0

1+2i  is a root

z= 1+2i

z^2 = (1+2i) (1+2i)

       = 1 +2i+2i +4i^2

       = 1 +4i -4

       = -3+4i

= (-1+5i) (1+2i)

-1+5i-2i+10i^2

  -1+3i-10

  -11+3i                

z^2 + (-1 + 5·i)·z + 14 - 7·i = 0

-3+4i  + -11+3i +14 - 7i

Combine like terms

-3-11 +14 +4i +3i-7i

0

So 1+2i is a root

1-2i  is not a root

z= 1-2i

z^2 = (1-2i) (1-2i)

       = 1 -2i-2i +4i^2

       = 1 -4i -4

       = -3-4i

= (-1+5i) (1-2i)

-1+5i+2i-10i^2

  -1+7i+10

  9+7i                

z^2 + (-1 + 5·i)·z + 14 - 7·i = 0

-3-4i  + 9+7i +14 - 7i

Combine like terms

-3+9 +14 -4i +7i-7i

20 -4i

So 1-2i is not a root

The complex conjugate being roots is only true for real coefficients

5,000

Step-by-step explanation:

the thousand's place in 4,712 is "4". You look at the number right after it, if it is less than 5, you don't change the thousands place, if it is equal or more than five you add 1 (so in this case 7 >= 5, so you add 1 to 4).