Multiplying square roots of negative numbers?

**MSUMathStdnt** · March 10th 2011, 09:13 PM

Which is right (and why)?

$17I+4-2 \cdot \sqrt{-2} \cdot \sqrt{-18} = 17I+4-2I \cdot \sqrt{2} \cdot I\sqrt{18}= \ldots = 16+17I$

or

$17I+4-2 \cdot \sqrt{-2} \cdot \sqrt{-18} = 17*I+4-2 \cdot \sqrt{36}= \ldots = -8+17I$

Actually, according to my calculator, the first one is right, so I really only want to know why.

Isn't $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ ? If so, then $\sqrt{-2} \cdot \sqrt{-18} = \sqrt{(-2)(-18)} = \sqrt{36}=6$ and the second one is "also" right.

Note that my calc says:
$\sqrt{(-2)(-18)}=6$ , and
$\sqrt{-2} \cdot \sqrt{-18}=-6$

Is the conclusion that $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ is only true for $a,b \ge 0$ ?

Thanks

**pickslides** · March 10th 2011, 11:01 PM

$\displaystyle \sqrt{-2}= \sqrt{-1\times 2}= \sqrt{-1}\times\sqrt{ 2}= i\times\sqrt{ 2}$

$\displaystyle \sqrt{-18}= \sqrt{-1\times 18}= \sqrt{-1}\times\sqrt{ 18}= i\times\sqrt{ 18}$

$\displaystyle \sqrt{-18}\times \sqrt{-2}= i\times\sqrt{ 2}\times i\times\sqrt{ 18}= i^2\times \sqrt{36}= -6$

**MSUMathStdnt** · March 11th 2011, 12:09 AM

Originally Posted by pickslides

$\displaystyle \sqrt{-2}= \sqrt{-1\times 2}= \sqrt{-1}\times\sqrt{ 2}= i\times\sqrt{ 2}$

$\displaystyle \sqrt{-18}= \sqrt{-1\times 18}= \sqrt{-1}\times\sqrt{ 18}= i\times\sqrt{ 18}$

$\displaystyle \sqrt{-18}\times \sqrt{-2}= i\times\sqrt{ 2}\times i\times\sqrt{ 18}= i^2\times \sqrt{36}= -6$

OK, I have no problem with that. But why is the rule $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ being violated when $a=-2$ and $b=-18$ ?

In other words, what's wrong with (the second one of) these:
$\sqrt{-1} \cdot \sqrt{-1} = \imath \cdot \imath=-1$
$\sqrt{-1} \cdot \sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1$

**Unknown008** · March 11th 2011, 12:14 AM

You are right, that rule is not always true.

Here is a wiki article part about it:

Square root - Wikipedia, the free encyclopedia

**MSUMathStdnt** · March 11th 2011, 12:54 AM

Originally Posted by Unknown008

You are right, that rule is not always true.

Here is a wiki article part about it:

Square root - Wikipedia, the free encyclopedia

I have been teaching my Algebra I students that rule for two years now. I can't decide whether I'm upset at the answer or relieved to have the answer.
Thanks for the link.

**Unknown008** · March 11th 2011, 01:17 AM

If they haven't learned about complex numbers yet, it's there's still some time

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