1.(x-1)(x-3)(x-4)(x-12)+5x^2=0 2.設複數z滿足z^2=8+6i,試求z的值.?

2016-10-27 2:08 pm

回答 (1)

2016-10-27 11:52 pm
✔ 最佳答案
1.(x-1)(x-3)(x-4)(x-12)+5x^2=0
Sol
(x-1)(x-3)(x-4)(x-12)+5x^2=0
(x-1)(x-12)(x-3)(x-4)+5x^2=0
(x^2+12-13x)(x^2+12-7x)+5x^2=0
(x^2+12)^2-20x(x^2+12)+91x^2+5x^2=0
(x^2+12)-20x(x^2+12)+96x^2=0
(x^2+12-8x)(x^2+12-12x)=0
(x-2)(x-6)(x^2-12x+12)=0
x=2 or x=6 or x=[12+/-√(144-48)]/2
x=2 or x=6 or x=6+/-2√6
2.設複數z滿足z^2=8+6i,試求z的值.?
Sol
設z=a+bi,a,b為實數
z^2=a^2-b^2+2abi
a^2-b^2=8
ab=3
b=3/a
8=a^2-b^2=a^2-9/a^2
8a^2=a^4-9
a^4-8a^2-9=0
(a^2-9)(a^2+1)=0
a^2=9 or a^2=-1(不合)
a=+/-3
(1) a=3
b=1
z=3+i
(2) a=-3
b=-1
z=-3-i


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