Can a number have 2 square roots? If you say no, surely 4 has 2 squared and minus 2 squared bothe as square roots? Do minus numbers get involved in square roots? And surely a minus number can't have a square root?
It's not exactly a 13+ question (which is what I'm doing), it's just out of curiosity.
But when you are restricting to real numbers it is standard to define the square root on b to be the positive real number a such that . In that sense a number has at most one square root. 2 is the square root of 4. (The reason we say that the two roots of are is precisely because itself is only the positive root.) We do this to make the square root function "single valued"- a very useful property.
Unfortunately, if we require functions of complex numbers be "single valued" we would lose most of our most useful functions! So for functions of complex variables we drop that requirement, making Plato's answer correct.
..after you find the positive root (called the principal square root) multiply it by -1 to get the negative root.
example: square root of 9 = 3 (3 is the principal square root) ; 3 x -1 = -3 (-3 is the negative root) Thus the square root of 9 has two answers: 3 and -3
Now, for the square root of a negative number, find the square root of the same positive number and multiply by the letter "i" to find your answer. Where "i" will mean your answer is an imaginary number.
example: square root of -9 = (square root of 9)x(i) =3i ; the negative root will be 3i x -1 = -3i Thus the square root of -9 has two answers: 3i and -3i
Now, for the square root of an imaginary number I usually use a square root calculator that can handle imaginary numbers or complex numbers. http://www.squarerootcalcultor.co can help.
example: square root of 9i = 2.12 + 2.12i (which is a complex number) ; the negative root is -2.12 - 2.12i Thus the square root of 9i has two answers: 2.12 + 2.12i and -2.12 - 2.12i
I hope that helps.
The symbol (read "the square root of x") is defined as a function. That is, for each x, there is only one . However, for any positive real number x, there are exactly 2 real numbers r1,r2 such that r12 = x and r22 = x. (In this case r1 = -r2.)
If we include complex numbers we can extend this concept and assert: for any complex number x, there are exactly n complex numbers r1,r2,...,rn such that rin = x, 1 ≤ i ≤ n. The roots ri may not be distinct, but always come in pairs of complex conjugates. (If n is odd, there will be at least one real ri, and the rest are the pairs of complex conjugates. All pairs of real numbers are complex conjugates.)
24 = 16,
(-2)4 = 16,
(2i)4 = 16,
(-2i)4 = 16,
Note: we define i = i2 = -1 (i2)2 = i4 = 1
(i is not a real number.)
defining i = √(-1) is somewhat problematic. the reason being, we don't know how to distinguish between i and -i (algebraically a+ib and a-ib act identitically: the map
z-->z* is a field automorphism).
the best we can say is i is "a" square root of -1 (in other words, orientation in the complex plane is chosen arbitrarily). behold the following "proof":
1 = √(1) = √((-1)(-1)) = √(-1)√(-1) = (i)(i) = -1,
which highlights the risky business of dealing with complex square roots. what happens is: multiplication in the complex plane can "rotate by more than a whole circle", the exponential function is no longer 1-1. so things like principal values and branches become very important (if we regard the complex number 1 = 1+0i as having angle 2π, it's "principal square root" is -1, if we regard it as having angle 0, it's principal square root is 1, the "standard convention").
perhaps the least confusing way to answer the OP's question is this:
for any real number a:
if a > 0, there are two real numbers that satisfy x2 = a. we call the positive one √a, and the negative one -√a.
if a = 0, there is only one number satisfying x2 = 0, namely: 0.
if a < 0, there are two complex numbers satisfying x2 = a, either one of which can be called the/a square root (there's no sense of "positiveness" in the complex plane).