complex number helnp me plz
回答 (3)
2010-02-18 6:12 am
✔ 最佳答案
x^2 + ax + b = 0For x = 3 + 4j is a root, then its conjugate x = 3 - 4j must be another root.
Sum of roots, -a = (3 + 4j) + (3 - 4j) = 6
So, a = -6
Product of roots, b = (3 + 4j)(3 - 4j) = (3)^2 - (4j)^2 = 9 - (-16) = 25
So, a = -6, b = 25
參考: Physics king
2010-02-18 6:11 pm
if x=3+4j is a root of x^2+ax+b=0，find the real numbers a and b
Sol
x=3+4j
x－3=4j
x^2－6x+9=－16
x^2－6x+25=0
a=－6，b=25
Sol
x=3+4j
x－3=4j
x^2－6x+9=－16
x^2－6x+25=0
a=－6，b=25
2010-02-18 6:22 am
Sum of roots, -a = (3 + 4j) + (3 - 4j) = 6
點解a會變左-a?
2010-02-17 22:30:33 補充：
Sum of roots, -a = (3 + 4j) + (3 - 4j) = 6
點解a會變左-a?
點解a會變左-a?
2010-02-17 22:30:33 補充：
Sum of roots, -a = (3 + 4j) + (3 - 4j) = 6
點解a會變左-a?
收錄日期: 2021-04-19 21:21:26
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