2007-08-16 16:45:59 · 9 answers · asked by phoenixfighter00 2 in Science & Mathematics ➔ Mathematics
how does the i=√-1 play into the whole equation?
2007-08-16 16:57:57 · update #1
Noting that i ² = - 1 :-
6 i + 15 i ² = x + 6 i
6 i - 15 = x + 6 i
x = - 15
2007-08-16 22:29:38 · answer #1 · answered by Como 7 · 2⤊ 0⤋
3i(2 + 5i) = x + 6i
6i + (5 * 3)(i^2) = x + 6i
since i = sqrt(-1), i^2 = -1, so
6i + (5 + 3)(i^2)
6i + 15(-1)
6i - 15
this is the same as saying
-15 + 6i, so x + 6i = -15 + 6i
x = -15
2007-08-16 18:06:00 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
i can be treated like any other variable, except that i^2= -1 (since [rt(x)]^2 = x). Your final answer should be in the form x=a+bi where a and b are real numbers.
3i(2+5i)=x+6i
6i+15i^2 = x+6i
x= -15
2007-08-16 16:52:33 · answer #3 · answered by Anonymous · 0⤊ 0⤋
hang on ..
just multiply it out:
3i(2+5i)=x+6i
6i + 15i^2 = x + 6i
6i - 15 = x + 6i
-15 = x
looks right
2007-08-16 16:49:18 · answer #4 · answered by atheistforthebirthofjesus 6 · 0⤊ 0⤋
First distribute....
6i +15i^2=x+6i
okay now i*i or i^2 = -1
so substitute it in....
6i + (15)(-1) = x + 6i
now subtract 6i off both sides (notice they cancel)
-15=x
2007-08-16 16:50:29 · answer #5 · answered by Anonymous · 0⤊ 0⤋
so 6i - 15 = x + 6i
therefore x = -15
2007-08-16 16:50:11 · answer #6 · answered by Anonymous · 0⤊ 0⤋
6i-15 = x+6i
x = -15
2007-08-16 16:49:45 · answer #7 · answered by sahsjing 7 · 2⤊ 0⤋
x + 6i = 6i - 15
x = - 15
2007-08-16 16:49:49 · answer #8 · answered by Helmut 7 · 0⤊ 0⤋
beats me...i'm not in skool yet:P
2007-08-16 16:52:58 · answer #9 · answered by A-Rod 2 · 0⤊ 0⤋