2006-12-29 12:52:02 · 7 answers · asked by wicked 1 in Science & Mathematics ➔ Mathematics
i=-1
(√9*16)i=
√144i=
12i=-12
2006-12-30 03:19:38 · answer #1 · answered by Anonymous · 4⤊ 0⤋
A square root other than zero always has two values.
sqrt(-9)*sqrt(-16) = sqrt{(-9)(-16)} = sqrt(144) = +12 or -12.
2006-12-29 13:29:04 · answer #2 · answered by ninasgramma 7 · 0⤊ 3⤋
It isn't all that easy. Square root of negative numbers is totally imaginary, as a number, positive or negative, can never yield a negative result when multiplied by itself. -9 and -16 do not have square roots. And I'm pretty sure that two such numbers multiplied with each other will never yield a real number. Such numbers are called complex numbers.
2006-12-29 13:06:40 · answer #3 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 3⤋
sqrt(-9) = 3i or -3i
sqrt(-16) = 4i or -4i
3i * 4i = 12i^2 = 12(-1) = -12
-3i * -4i = 12i^2 = -12
-3i * 4i = -12i^2 = -12(-1) = 12
3i * -4i = -12i^2 = 12
ANS : 12 or -12
2006-12-29 14:20:16 · answer #4 · answered by Sherman81 6 · 3⤊ 0⤋
square root of 9 is 3
square root of 16 is 4
3x4=12
2006-12-29 12:59:50 · answer #5 · answered by Each1Teach1 3 · 0⤊ 4⤋
Recall that a^n * b^n = (ab)^n
sqrt(-9) = -9^.5
sqrt(-16) = -16^.5
-9^.5 * -16^.5
= (-9 * -16)^.5
= 144^.5
=12
2006-12-29 12:58:40 · answer #6 · answered by Tim P. 5 · 1⤊ 3⤋
sqrt(-1)=i
-1=i^2
so,
sqrt(-9)*Sqrt(-16)=
3i*4i
12*i^2
-12
Answer:-12
2006-12-29 13:03:05 · answer #7 · answered by Anonymous · 5⤊ 0⤋