3

For example : $\sqrt {-18} \times \sqrt {-12}$ Would I start by multiplying the 2 numbers under a square root, In which case the double negatives cancel out? $ \sqrt {-18\times-12} = \sqrt {216} = 6\sqrt {6}$

Or get the i out of the square roots in which case I get a $i^2$?
$$ \sqrt {18}i \times \sqrt {12}i = 6\sqrt {6}i^2 = - 6\sqrt {6}$$

Kareem Atef
  • 31
  • 1
  • 4
  • 4
    You don't really “solve” expressions. Do you mean simplify? And what you're discovering is that the square root rules can't be extended to complex numbers that way. – Matthew Leingang Oct 20 '17 at 14:21
  • 1
    Have a read of this post. It explains why you cannot take negative numbers into square roots – John Doe Oct 20 '17 at 14:21
  • 1
    I know this problem has been discussed on this site before, but in a nutshell, the properties for radicals do not blindly apply if the insides are negative. By the way, your second answer is right – imranfat Oct 20 '17 at 14:21
  • 2
    When dealing with values other than non-negative reals, it isn't generally true that $\sqrt{a}\sqrt{b}=\sqrt{ab}$. – Thomas Andrews Oct 20 '17 at 14:24
  • Thanks for the quick answer! – Kareem Atef Oct 20 '17 at 14:26

2 Answers2

1

Simply

$$i\sqrt{18}\cdot i\sqrt{12}$$

$$-1\sqrt{12\cdot 18}$$

$$-1\sqrt{216}$$

$$-6\sqrt{6}$$

ALWAYS start with $i$

MCCCS
  • 1,625
0

I always emphasize to not use square roots on negative numbers! Even though one can extend the function $\sqrt\cdot$ to act like

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

for positive real numbers $x$, there is some random choice here: why $i\sqrt x$ and not $-i\sqrt x$? All arithmetic operations are absolutely invariant under an exchange $i\leftrightarrow-i$. So our choice is arbitrary and this is not good. Note that the same symmetry does not exist between $1$ and $-1$ and so it is absoloutely okay to define $\sqrt 4=2$ and not $-2$.

Yes I know, we learn that complex numbers are there to compute square roots of negative number, but this is more meant for applications like solving $x^2=-1$, which has well defined solutions $\{i,-i\}$. As I said, one can define the square root function $\sqrt\cdot$ to give $\sqrt{-1}=i$, but then we have to live with the fact that we cannot use our beautiful power rules!

M. Winter
  • 29,928