Evaluate the square root of -19 multiplied by the square root of -19.

Step-by-step solution

1. 1

   Do not use the product rule for radicals here. The identity $\sqrt a\cdot\sqrt b=\sqrt{ab}$ is only valid when $a,b\ge0$. Applying it blindly would give $\sqrt{(-19)(-19)}=\sqrt{361}=19$ - the wrong sign, and the classic trap in this problem.

2. 2

   Convert to the imaginary unit first. The principal square root of a negative number is defined through $i=\sqrt{-1}$:

   $\sqrt{-19}=i\sqrt{19}$

   Once every radicand is non-negative, ordinary algebra is safe again.

3. 3

   Multiply the two forms.

   $\left(i\sqrt{19}\right)\left(i\sqrt{19}\right)=i^{2}\left(\sqrt{19}\right)^{2}=i^{2}\cdot 19$

4. 4

   Apply $i^2=-1$.

   $i^{2}\cdot 19=-1\cdot 19=-19$

5. 5

   Confirm with the definition of a square root. For any number $z$, $\left(\sqrt z\right)^2=z$ by definition - so $\left(\sqrt{-19}\right)^2$ must be $-19$. This matches the computed result and shows why $+19$ could never be right ✓. A numerical check with $\sqrt{19}=4.3589$: $(4.3589i)^2=-19.000$ ✓.