How Do You Find the Square Root of a Negative Number?

A negative number cannot be squared to produce a negative result if you restrict yourself to real numbers. To make square roots of negative numbers possible, mathematics extends the real numbers to the complex numbers, which include the imaginary unit $i$, defined by $i^2=-1$.

This lets you express the square root of a negative real number using an ordinary positive square root. There is one detail to keep straight: the radical symbol usually means the principal square root, while a quadratic equation such as $z^2=-9$ has two solutions.

Why negative numbers have no real square roots

For every real number $x$, its square is nonnegative: $x^2\geq 0$. For example, $5^2=25$ and $(-5)^2=25$. So there is no real number whose square is $-25$.

The complex numbers address this by including $i$, where $i^2=-1$. For any positive real number $a$, the number $i\sqrt{a}$ squares to $-a$:

$$(i\sqrt{a})^2=i^2(\sqrt{a})^2=-1\cdot a=-a.$$

Thus, using the usual principal-root convention,

$$\sqrt{-a}=i\sqrt{a}\qquad (a>0).$$

Work through an example

Find the principal square root of $-36$. First write $-36=(-1)\cdot36$. Then use $i=\sqrt{-1}$ and simplify the positive square root:

$$\sqrt{-36}=\sqrt{-1}\sqrt{36}=i\cdot6=6i.$$

Check by squaring: $(6i)^2=36i^2=-36$. So the principal square root is $6i$.

The same reasoning gives $\sqrt{-7}=i\sqrt{7}$. It is usually left in that form because $\sqrt{7}$ is not an integer.

The radical and the equation have different answers

In the complex numbers, every nonzero number has two square roots. The symbol $\sqrt{z}$ conventionally denotes just the principal square root. For a negative real number $-a$ with $a>0$, this is $i\sqrt{a}$. But solving the equation $z^2=-a$ gives both roots:

$$z=\pm i\sqrt{a}.$$

For example, $\sqrt{-9}=3i$ means that the principal root is $3i$. The equation $z^2=-9$ has two solutions, $z=3i$ and $z=-3i$, since both square to $-9$.

This distinction is the same as for positive numbers: the principal value $\sqrt{9}$ is $3$, whereas the equation $x^2=9$ has solutions $x=3$ and $x=-3$. The radical does not list every solution to a square equation.

Why the real-number rules need care

It is tempting to use $\sqrt{ab}=\sqrt{a}\sqrt{b}$ without checking the inputs. That rule works for nonnegative real $a$ and $b$, but it does not hold for all complex numbers when square roots mean principal roots. For instance,

$$\sqrt{(-1)(-1)}=\sqrt{1}=1,$$

but

$$\sqrt{-1}\sqrt{-1}=i\cdot i=-1.$$

The two sides differ. When simplifying a negative real under a square root, a reliable approach is to factor out exactly one $-1$ and use $i=\sqrt{-1}$, as in $\sqrt{-36}=i\sqrt{36}$. Avoid splitting a product into radicals if that would put negative or complex numbers into the factors.

Common mistakes

• Saying there is no square root at all: There is no real square root of a negative real number, but there are complex square roots.
• Giving only one solution to an equation: $\sqrt{-a}$ is the principal root; $z^2=-a$ has the two solutions $\pm i\sqrt{a}$.
• Forgetting that $i^2=-1$: $i$ is not an ordinary real number, and its square is negative one, not one.
• Applying real radical rules to complex values: In particular, do not assume $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for arbitrary complex $a$ and $b$.

If a problem explicitly asks for a real square root, say that none exists for a negative input. If it allows complex numbers, use $i\sqrt{a}$ for the principal root of $-a$, or $\pm i\sqrt{a}$ when solving the equation $z^2=-a$.