Please solve ques no 3?
回答 (2)
2017-10-17 10:25 pm
3.
a, b, c are the roots of x³ - 3x² + 6x + 1 = 0, then
a + b + c = 3
ab + bc + ca = 6
abc = -1
(ab + bc + ca)/3
= 6/3
= 2
[(1/bc) + (1/ca) + (1/ab)]/3
= [(a/abc) + (b/abc) + (c/abc)]/3
= [(a + b + c)/abc]/3
= [3/(-1)]/3
= -1
The coordinates of the centroid
= ((ab + bc + ca)/3, [(1/bc) + (1/ca) + (1/ab)]/3)
= (2, -1)
a, b, c are the roots of x³ - 3x² + 6x + 1 = 0, then
a + b + c = 3
ab + bc + ca = 6
abc = -1
(ab + bc + ca)/3
= 6/3
= 2
[(1/bc) + (1/ca) + (1/ab)]/3
= [(a/abc) + (b/abc) + (c/abc)]/3
= [(a + b + c)/abc]/3
= [3/(-1)]/3
= -1
The coordinates of the centroid
= ((ab + bc + ca)/3, [(1/bc) + (1/ca) + (1/ab)]/3)
= (2, -1)
2017-10-17 10:19 pm
x^3-3x^2+6x+1 = 0
http://www.wolframalpha.com/input/?i=graph+solve+x%5E3-3x%5E2%2B6x%2B1+%3D+0
There is only one real root. The other two are complex.
http://www.wolframalpha.com/input/?i=graph+solve+x%5E3-3x%5E2%2B6x%2B1+%3D+0
There is only one real root. The other two are complex.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171017140500AAmyKg1