The square root of (-x) is the value that, when multiplied by itself, gives (-x). On the flip side, in the real number system, (\sqrt{-x}) is real only when (-x) is nonnegative, meaning when (x \leq 0). ” The answer depends on what kind of numbers we are using. That said, in simple terms, (\sqrt{-x}) asks: “What number squared equals (-x)? In the complex number system, (\sqrt{-x}) can be expressed using the imaginary unit (i), where (i^2 = -1).
Introduction to (\sqrt{-x})
The expression (\sqrt{-x}) can look confusing because the variable (x) is hidden inside a negative sign. To understand it, we need to ask what values (x) is allowed to take and whether we are working with real numbers or complex numbers.
The square root operation reverses squaring. For example:
[ 3^2 = 9 ]
so
[ \sqrt{9} = 3 ]
using the principal square root. Even so, both (3) and (-3) square to (9), because:
[ (-3)^2 = 9 ]
The symbol (\sqrt{9}) usually means the principal square root, which is the nonnegative root. So (\sqrt{9} = 3), not (\pm 3). When solving equations, however, we often write (\pm\sqrt{9}) Worth keeping that in mind..
For (\sqrt{-x}), the key question is whether (-x) is positive, negative, or zero.
If (x) is positive
Suppose (x) is a positive number, such as (x = 4). Then:
[ -x = -4 ]
So:
[ \sqrt{-x} = \sqrt{-4} ]
In the real number system, there is no real number whose square is (-4). Squaring any real number gives a nonnegative result:
[ 2^2 = 4 ]
[ (-2)^2 = 4 ]
[ 0^2 = 0 ]
No real number squared gives a negative number. Which means, if (x > 0), then (\sqrt{-x}) is not a real number The details matter here..
In the complex number system, we introduce the imaginary unit (i), defined as:
[ i = \sqrt{-1} ]
This means:
[ i^2 = -1 ]
Using this, we can write:
[ \sqrt{-4} = \sqrt{4 \cdot -1} = \sqrt{4}\sqrt{-1} = 2i ]
So if (x = 4), then:
[ \sqrt{-x} = 2i ]
More generally, if (x) is positive, then:
[ \sqrt{-x} = i\sqrt{x} ]
For example:
[ \sqrt{-9} = 3i ]
because:
[ (3i)^2 = 9i^2 = 9(-1) = -9 ]
If (x) is zero
If (x = 0), then:
[ -x = 0 ]
So:
[ \sqrt{-x} = \sqrt{0} = 0 ]
This is both real and complex, and it is its own square root because:
[ 0^2 = 0 ]
So when (x = 0), the square root of (-x) is simply:
[ 0 ]
If (x) is negative
Suppose (x = -4). Then:
[ -x = -(-4) = 4 ]
So:
[ \sqrt{-x} = \sqrt{4} = 2 ]
In this case, (\sqrt{-x}) is a real number. This happens because a negative sign in front of a negative number makes the value positive.
So if (x < 0), then (-x > 0), and:
[ \sqrt{-x} ]
is a real number.
For example:
[ x = -1 ]
[ \sqrt{-x} = \sqrt{-(-1)} = \sqrt{1} = 1 ]
[ x = -25 ]
[ \sqrt{-x} = \sqrt{25} = 5 ]
When (x) is negative, (\sqrt{-x}) is the same as (\sqrt{|x|}), because (-x) equals the absolute value of (x) Nothing fancy..
The real-number answer
If we are working only with real numbers, then (\sqrt{-x}) is defined only when the expression inside the square root is nonnegative.
That means:
[ -x \geq 0 ]
Solving this inequality:
[ x \leq 0 ]
So, in the real number system:
[ \sqrt{-x} ]
is real only when:
[ x \leq 0 ]
If (x > 0), then (\sqrt{-x}) is not real Simple as that..
The complex-number answer
If complex numbers are allowed, then (\sqrt{-x}) can be written for most values
