SOLUTION: One root of {{{x^4 - 3x^2 - 4 = 0}}} is 'i'. What are the other three roots?

Algebra ->  Square-cubic-other-roots -> SOLUTION: One root of {{{x^4 - 3x^2 - 4 = 0}}} is 'i'. What are the other three roots?      Log On


   

Question 1102277: One root of x%5E4+-+3x%5E2+-+4+=+0 is 'i'. What are the other three roots?
Found 2 solutions by ikleyn, josgarithmetic:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Factor

x%5E4+-+3x%5E2+-+4 = %28x%5E2-4%29%2A%28x%5E2%2B1%29.


The polynomial  x^2 - 4 has the real roots  2  and  -2.


The polynomial x^2+1 has complex roots  i  and  -i.


The list of all roots is  -2,  2,  i  and  -i.


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I do not need your hint.


Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Possible rational roots to try:  -4, -2, -1, 1, 2, 4

-1   |    1   0    -3   0    -4
     |        -1    1   2    -2
     ---------------------------------
         1   -1    -2   2     -6


-2   |    1   0    -3   0    -4
     |       -2     4   -2    4
     ---------------------------------
         1   -2    1    -2   0        -2 is a root, (x+2)

2     |    1   -2    1    -2
      |         2    0     2
      ----------------------------
           1    0    1     0          2 is a root, (x-2)

Other roots are found using x%5E2%2B1=0, from the result of the last synthetic division shown.

Note the use of 'i' in i^2=-1, imaginary number i.
i=-sqrt%28-1%29 or i=sqrt%28-1%29

x%5E2%2B1=0
x%5E2=-1
system%28x=-sqrt%28-1%29%2Cor%2Cx=sqrt%281%29%29
system%28x=-i%2Cor%2Cx=i%29



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ALL THE ROOTS:
system%28-2%2C2%2C-i%2Ci%29
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