would you use FOIL when you are multiplying imaginary #'s?
like (2+7i) (4-2i) or what do you do?
2006-10-17 12:55:18 · 12 answers · asked by Mia16 3 in Science & Mathematics ➔ Mathematics
Yes, exactly.
First: 2 * 4 = 8
Outer: 2 * -2i = - 4i
Inner: 7i * 4 = 28i
Last: 7i * -2i = -14 * i² = -14 * -1 = 14
Watch the last part were i² = -1.
Now put it back together:
8 - 4i + 28i + 14
Group the real parts and the imaginary parts:
= (8 + 14) + (28i - 4i)
Simplify:
= 22 + 24i
2006-10-17 12:57:27 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
First multiply the first number in the first set (2) with both numbers in the second set (4-2i) now multiply 2(4-2i) = 2(4)-2(2i)= 8-4i. Now do the same thing only with the second number (7i) ,
7i(4-2i)=7i(4)-7i(2i)=28i-14i^2 (Remember i^2 = -1) so 28i-14(-1)= 28i+14 now collect like terms 8-4i+28i+14 combine the whole numbers with other whole numbers and the imaginary with the imaginary. So 8+14 , -4i+28i = 22+24i That's your answer
2006-10-17 13:11:09 · answer #2 · answered by dark&pure? 3 · 0⤊ 0⤋
Foiling with imaginaries. Fun.
Basically the same as normal, but with a few twists, so watch closely.
First: 2X4, which is 8
Outside: 2X-2i, which is -4i
Inside: 7iX4, which is 28i
Last: 7iX-2i, which is -14i2 (i squared)
Then add and simplify: 8-4i+28i-14i2 (i squared)
i2 (i squared) is -1, so -14i2 would be 14
Thus: 8+14=22, 28i-4i=24i, and your answer is 24i+22.
Get all of that?
2006-10-17 13:03:25 · answer #3 · answered by eeaghk2112 2 · 0⤊ 0⤋
definite, there's a reason. really to be ready to precise the suggestions of equations like x^2 + a million = 0, which has no authentic suggestions. Equations and formula that bring about imaginary numbers arise in many situations in the sciences and engineering. Voltage, cutting-edge, and means, as an get mutually, are countless the parts that is expressed in words of imaginary numbers.
2016-12-04 22:46:08 · answer #4 · answered by stanberry 4 · 0⤊ 0⤋
(2 + 7i)x(4 -2i) = 8 -4i +28i -14i^2
= 8 + 24i -14(-1)
= 8 + 14 + 24i
= 22 + 24i
2006-10-17 13:15:17 · answer #5 · answered by quark_sa 2 · 0⤊ 0⤋
ahhh imaginary numbers. Well use foil your correct then work the rest of the problem out so hence you get 8+24i-14i^2 and as you know i^2 is a actual value (-1) so its 8+24i-(-14) = 22 +24i i think
2006-10-17 12:58:44 · answer #6 · answered by gordon_benbow 4 · 0⤊ 0⤋
yes you do use foil. For example the answer to that problem woul dbe 8-4i +28i-14isquared. Recall that i squared = -1
22+24i would be my guess as to the answer
2006-10-17 12:58:44 · answer #7 · answered by asd589 2 · 0⤊ 0⤋
yes,you may.first you multiply 2 by 4,then by -2i.second you multiply 7i by 4,& then by -2i.finally you group them together &you will get the answer.
2006-10-17 13:07:31 · answer #8 · answered by Gardenia 6 · 0⤊ 0⤋
yes, when you get i^2, you simply call that -1
2006-10-17 12:59:36 · answer #9 · answered by zmonte 3 · 0⤊ 0⤋
yes
2006-10-17 12:59:23 · answer #10 · answered by Anonymous · 0⤊ 0⤋