• Support PF! Buy your school textbooks, materials and every day products Here!

Roots of a squared polynomial ( complex numbers)

  • Thread starter Axe199
  • Start date
  • #1
48
0

Homework Statement



problem in a pic attached

Homework Equations





The Attempt at a Solution


i solved i and ii a , when it came to b , i just said that every one of the 3 roots will be squared having 2 roots 1 + and 1 - but then i read the marking schemes ( also attached) , and i got really confused and now i dont know what he wants from me
 

Attachments

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement



problem in a pic attached

Homework Equations





The Attempt at a Solution


i solved i and ii a , when it came to b , i just said that every one of the 3 roots will be squared having 2 roots 1 + and 1 - but then i read the marking schemes ( also attached) , and i got really confused and now i dont know what he wants from me
He wants you to tell him what are all the square roots of -2, -2+2*sqrt(3)*i and -2 - 2*sqrt(3)*i.
 
  • #3
48
0
He wants you to tell him what are all the square roots of -2, -2+2*sqrt(3)*i and -2 - 2*sqrt(3)*i.
why cant i just say (z+2)^2= 0 and solve and do the same for the complex roots?
why do i have to find the roots of the -2?
 
  • #4
1,013
70
why cant i just say (z+2)^2= 0 and solve and do the same for the complex roots?
why do i have to find the roots of the -2?
(z + 2)2 is not a factor of p(z2). It is a factor of (p(z))2, however, a completely different polynomial.
The polynomial is not the object being squared. p(z2) only means that the variable z is being replaced with the variable z2. So, as a result, we know that z2 + 2 is a factor.
 
  • Like
Likes 1 person
  • #5
48
0
(z + 2)2 is not a factor of p(z2). It is a factor of (p(z))2, however, a completely different polynomial.
The polynomial is not the object being squared. p(z2) only means that the variable z is being replaced with the variable z2. So, as a result, we know that z2 + 2 is a factor.
okay...thanks very much, tht was a tricky one, or maybe i am stupid :D
either way , thanks , i got it
 

Related Threads on Roots of a squared polynomial ( complex numbers)

  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
7
Views
9K
  • Last Post
Replies
3
Views
2K
Replies
3
Views
1K
Replies
3
Views
13K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
1K
Top