Square roots of negative real numbers

Where $i$ comes from. Real numbers cannot supply a square root of a negative number, since any real number squared is zero or positive. The imaginary unit $i$ is introduced precisely to fill this gap, defined by $i^2=-1$ (equivalently $i=\sqrt{-1}$). Once $i$ exists, a square root can be assigned to every negative number, not just $-1$.

The defining rule. For any positive real number $k$,

where: $k$ is a positive real number, $\sqrt{k}$ is its ordinary (non-negative) real square root, and $i$ is the imaginary unit with $i^2=-1$. Squaring the right-hand side checks it out immediately: $(i\sqrt{k})^2=i^2\cdot k=-k$, confirming $i\sqrt{k}$ really is a square root of $-k$.

Two roots, one convention. Every nonzero number has exactly two square roots, differing only in sign. For $-k$ these are $i\sqrt{k}$ and $-i\sqrt{k}$, since $(-i\sqrt{k})^2=i^2k=-k$ too — exactly parallel to how a positive real $k$ has two real square roots, $\sqrt{k}$ and $-\sqrt{k}$. By convention, the symbol $\sqrt{-k}$ itself always names the principal value $i\sqrt{k}$, the same convention already used for positive reals, where $\sqrt{k}$ means the non-negative root, not $\pm\sqrt{k}$.

Where the real-number rule breaks. The identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$ holds whenever at least one of $a,b$ is non-negative, but fails when both are negative. Multiplying radicands first, $\sqrt{-1}\cdot\sqrt{-1}\overset{\text{wrong}}{=}\sqrt{(-1)(-1)}=\sqrt{1}=1$, but the true value, found by converting to $i$-form first, is $\sqrt{-1}\cdot\sqrt{-1}=i\cdot i=i^2=-1$. The shortcut and the correct answer disagree in sign — a gap of exactly $2$, not a rounding error, so it can never be waved away.

Why the failure happens. $\sqrt{a}\sqrt{b}=\sqrt{ab}$ is a theorem about real square roots; it was never proved, and does not hold, for complex numbers written with the $\sqrt{\phantom{a}}$ symbol over a negative argument. The safe procedure sidesteps the whole issue: rewrite every $\sqrt{-k}$ as $i\sqrt{k}$ the moment it appears, and only then multiply, add, or combine using ordinary complex-number arithmetic (where $i^2=-1$ is applied explicitly, never assumed away).

Where this shows up immediately. Solving $ax^2+bx+c=0$ by the quadratic formula produces $\sqrt{D}$ where $D=b^2-4ac$; whenever $D<0$, that square root must be converted to $i$-form before anything else is done with it. For $x^2+2x+5=0$, $D=4-20=-16$, and the correct next step is $\sqrt{-16}=i\sqrt{16}=4i$ — never an attempt to manipulate $\sqrt{-16}$ as though it were an ordinary real radical.

Addition and subtraction are unaffected. The invalid-shortcut trouble is specific to multiplication; adding or subtracting square roots of negative numbers is safe once each term has been converted to $i$-form, because addition of complex numbers never mixes real and imaginary parts incorrectly the way the multiplication shortcut does. So $\sqrt{-4}-\sqrt{-1}=2i-i=i$ works exactly as expected, with no hidden sign trap — the caution in this section applies specifically to products of two negative-radicand roots, not sums.

Problem: Simplify $\sqrt{-16}+\sqrt{-9}$, and separately simplify $\sqrt{-3}\times\sqrt{-27}$, showing why writing $\sqrt{(-3)(-27)}$ directly would be wrong.

Step 1. Convert each square root to $i$-form using $\sqrt{-k}=i\sqrt{k}$: $\sqrt{-16}=i\sqrt{16}=4i$ and $\sqrt{-9}=i\sqrt{9}=3i$.

Step 2. Add the converted terms as ordinary complex numbers: $\sqrt{-16}+\sqrt{-9}=4i+3i=7i$.

Step 3. For the product, convert first: $\sqrt{-3}=i\sqrt{3}$ and $\sqrt{-27}=i\sqrt{27}=i\cdot3\sqrt{3}=3\sqrt{3}\,i$. Multiply: $\sqrt{-3}\times\sqrt{-27}=(i\sqrt{3})(3\sqrt{3}\,i)=3i^2(\sqrt{3}\cdot\sqrt{3})=3i^2(3)=9i^2=9(-1)=-9$.

Step 4. Contrast with the invalid shortcut: multiplying radicands directly gives $\sqrt{(-3)(-27)}=\sqrt{81}=9$, the wrong sign, matching the general result from the derivation above ($\sqrt{-p}\times\sqrt{-q}=-\sqrt{pq}$ with $p=3,q=27$, giving $-\sqrt{81}=-9$). Final answers: $\sqrt{-16}+\sqrt{-9}=7i$ and $\sqrt{-3}\times\sqrt{-27}=-9$.

Competitive stretch

Apply the same habit to the quadratic $x^2+2x+5=0$: the discriminant is $D=2^2-4(1)(5)=4-20=-16$, so $\sqrt{D}=\sqrt{-16}=i\sqrt{16}=4i$ is substituted into $x=\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{-2\pm4i}{2}=-1\pm2i$, giving the complex roots directly — no separate rule is needed once $\sqrt{-16}$ has been converted to $4i$ up front.

• Multiplying two negative-radicand square roots directly as $\sqrt{ab}$ instead of converting both to $i$-form first — this silently flips the sign of the final answer.
• Writing $\sqrt{-k}=-\sqrt{k}$, dropping the $i$ entirely, instead of the correct $i\sqrt{k}$; a negative real number is never a square root of another negative number here.
• Forgetting that $\sqrt{-k}$ has two square roots, $\pm i\sqrt{k}$; the bare symbol $\sqrt{-k}$ by convention always means the principal value $i\sqrt{k}$, not both.
• Losing track of $i^2=-1$ partway through a simplification and accidentally treating $i^2$ as $+1$, which reintroduces the very sign error the $i$-form conversion was meant to prevent.
• Assuming the multiplication trap also applies to addition or subtraction — it does not; adding $i$-form terms is ordinary complex-number addition and carries no sign surprise.

Ask the tutor: want a few more practice problems combining and multiplying square roots of negative numbers without sign slips, or a walkthrough connecting this straight into solving quadratics with negative discriminants?