How many complex roots does this equation have
0= 5x^6 + x^4 - 3
Also how many rational roots does this equation have
x^4 + 3x^2 + 2 = 0
how many roots do these have?
2016-04-25 6:12 pm
回答 (3)
2016-04-25 6:41 pm
5x^6 + x^4 - 3 = 0
Six complex roots; two of those are real and four are non-real.
http://www.wolframalpha.com/input/?i=0%3D+5x^6+%2B+x^4+-+3
x^4 + 3x^2 + 2 = 0
Four roots, but they're all imaginary, so no real roots, and therefore no rational roots.
http://www.wolframalpha.com/input/?i=x^4+%2B+3x^2+%2B+2+%3D+0
Six complex roots; two of those are real and four are non-real.
http://www.wolframalpha.com/input/?i=0%3D+5x^6+%2B+x^4+-+3
x^4 + 3x^2 + 2 = 0
Four roots, but they're all imaginary, so no real roots, and therefore no rational roots.
http://www.wolframalpha.com/input/?i=x^4+%2B+3x^2+%2B+2+%3D+0
2016-04-25 6:22 pm
Not sure how you can tell this without seeing a graph.
A 6th degree polynomial will have 6 roots. Those that aren't real are complex.
Looking at a graph of the curve, I see 2 real roots, so it then has 4 complex roots.
for the second one, it never crosses the x-axis, so there are 0 real roots and 4 complex roots.
A 6th degree polynomial will have 6 roots. Those that aren't real are complex.
Looking at a graph of the curve, I see 2 real roots, so it then has 4 complex roots.
for the second one, it never crosses the x-axis, so there are 0 real roots and 4 complex roots.
2016-04-25 7:23 pm
Graphing the first one is the easiest.
The second is a quadratic function in x^2, so
x^2 = (-3 ± √9 - 8)/2 = (-3 ± 1)/2 = -2 or -1
I guess that's easier to see from factoring, since the expression is
(x^2 + 1)(x^2 + 2)
In either event, the 4 solutions are complex, ±√-1, ±√-2
The second is a quadratic function in x^2, so
x^2 = (-3 ± √9 - 8)/2 = (-3 ± 1)/2 = -2 or -1
I guess that's easier to see from factoring, since the expression is
(x^2 + 1)(x^2 + 2)
In either event, the 4 solutions are complex, ±√-1, ±√-2
收錄日期: 2021-05-01 20:52:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160425101246AAdYvYi