Is it incorrect to factor a square root of a positive number into two negative roots?

Question

I was learning to simplify a square root by factoring within the radical in order to organize it more effectively, such as $\sqrt{-52} = \sqrt{-1}\sqrt{13}\sqrt{4} = 2(\sqrt{13})i$ ... however, I was told it was incorrect to factor a positive number with two negatives, such as $\sqrt{16} = \sqrt{-4}\sqrt{-4}$, as this will produce a nonsensical answer. However, I find it makes sense!

$\sqrt{16} = \sqrt{-4}\sqrt{-4}$ = $2i(2i)$ $= 4i^2 = -4$... and since $-4^2 = 16$, it is not nonsense, but just the opposite of the the principle root.

Was my tutor erroneous, or am I?... Aside from this being less simple than the standard way.

$a^2=b^2$ does not imply that $a=b$. Just because $(-4)^2=(\sqrt{16})^2$ this does not imply that $-4=\sqrt{16}$. Remember $-4\neq +4$. Be careful about factoring with multiple negatives inside, else you'll wind up with contradictions like $1=\sqrt{-1\times -1}=\sqrt{-1}\times\sqrt{-1}=-1$ — JMoravitz, Nov 03 '17 at 04:19
It is always true that $\sqrt{ab} = \pm \sqrt{a}\sqrt{b}$. In particular $ 4 =\sqrt{16} = - \sqrt{-4} \sqrt{-4} = (-2i) (2i)$ where as you see it is effectively the product of two complex square roots of $-4$. To see what is the correct sign in $\pm$, you need to fix some branch(es) of $\sqrt{...}$ and be careful. — reuns, Nov 03 '17 at 04:31

Jaideep Khare · Answer 1 · 2017-11-03T04:22:54.377

3

$\sqrt x$ is a single valued function which is always positive.

Therefore, the answer to $\sqrt {16}$ is $\color{red}{\text{only}\; 4}$.

What you're doing wrong is that

$$ \sqrt {a \times b} =\sqrt a \times \sqrt b \; \text{if} \; a,b >0$$

Hence, $$\sqrt{(-4) \times (-4)}\neq \sqrt{-4}\times \sqrt{-4}$$

edited Nov 03 '17 at 04:22

answered Nov 03 '17 at 04:16

Jaideep Khare

19,293

Is it incorrect to factor a square root of a positive number into two negative roots?

1 Answers1