2007-12-09 21:40:59 · 13 answers · asked by Christopher 2 in Science & Mathematics ➔ Mathematics
Square roots of negative numbers
Negative numbers have square roots that lie outside the real numbers: Multiplying a real number by itself always results in a positive number. The square roots of negative numbers involve what are called imaginary numbers:
which for example leads to
The statements in "Two square roots for each number" also apply to negative numbers
square root of - 4 = square root of 4 times square root of -1
square root of 4 = 2
square root of -1 = i
therefore, square root of - 4 = 2i
2007-12-09 21:54:22 · answer #1 · answered by detektibgapo 5 · 3⤊ 5⤋
Square Root Of Negative 4
2016-09-29 00:56:25 · answer #2 · answered by weafer 4 · 0⤊ 0⤋
2
2007-12-09 21:43:32 · answer #3 · answered by Anonymous · 0⤊ 6⤋
the square root of -4 is complex numbers wher i=root -1.
.: root -4 = 2i
2007-12-09 23:00:29 · answer #4 · answered by ....................... 3 · 0⤊ 2⤋
I dont think there is a sqare root to -4 because to get a negative it must be a positive number times a negative number ( ie. -2 and 2 ) But to be a square the numbers must be the same ( ie -2 * -2 = 4 and 2*2=4 but neither = -4 )
2007-12-09 21:46:49 · answer #5 · answered by JJ 3 · 0⤊ 6⤋
√(- 4) = √ (4 i ²) = 2 i (taking + ve square root)
2007-12-10 04:05:32 · answer #6 · answered by Como 7 · 4⤊ 1⤋
That is a trick question.
You can't get the square root of a negative number.
A number times itself is a square root.
A negative times a negative is a positive.
A positive times a positive is a positive.
2007-12-09 21:47:28 · answer #7 · answered by Dennis Fargo 5 · 1⤊ 6⤋
ohh dear, its a complex number/imaginary number..
the square root of negative four is 2i.. that's it!
2007-12-09 21:50:30 · answer #8 · answered by fur_elise 2 · 1⤊ 2⤋
+ and - 2i. i is defined as the sq root of -1
2007-12-09 21:45:57 · answer #9 · answered by Anonymous · 1⤊ 2⤋
underroot -4 = underroot4i=+/-2i,
where i=underroot -1
2007-12-09 21:48:32 · answer #10 · answered by Abhinav 1 · 0⤊ 3⤋