How do you simplify #5i sqrt(-54)# ?

3 Answers
oaikitty
Sep 27, 2017

#-15sqrt6#

Explanation:

#5ixxsqrt(-1)xxsqrt(54)#

#5ixxixxsqrt54#

#5i^2xxsqrt54#

#5(-1)xxsqrt54#

#-5xxsqrt54#

#-5xxsqrt9sqrt6#

#-5xx3sqrt6#

#-15sqrt6#

Somebody N.
Sep 27, 2017

#-15sqrt(6)#

Explanation:

#5isqrt(-54) => 5isqrt(6xx9xx-1)#

We can take out the root of 9:

#5ixx3sqrt(6xx-1)=> 15isqrt(6xx-1)=> 15isqrt(6)sqrt(-1)#

#sqrt(-1) = i#

So we have:

#15ixxisqrt(6)#

#ixxi => i^2= -1#

So this gives:

#-15sqrt(6)#

George C.
Sep 27, 2017

#5isqrt(-54) = -15sqrt(6)#

Explanation:

Why another answer?

Because you should know that it is easy to make errors when it comes to square roots of negative (and complex) numbers.

The problem is that every non-zero number has two square roots, and the choice between them is a little arbitrary.

To see that there is a potential problem, consider the common "rule":

#sqrt(ab) = sqrt(a)sqrt(b)#

then note that:

#1 = sqrt(1) = sqrt(-1 * -1) != sqrt(-1) * sqrt(-1) = -1#

Ouch! The "rule" breaks if #a < 0# and #b < 0#.

Let's tread a little more carefully...

We use the symbol #sqrt# to denote the principal square root.

If #n > 0# then its principal square root is the positive one.

If #n < 0# then by convention, its principal square root is:

#sqrt(n) = i sqrt(-n)#

where #i# is the imaginary unit, satisfying #i^2=-1#

With these conventions, we can safely state:

#sqrt(ab) = sqrt(a)sqrt(b)" if "a >= 0" or "b >= 0#

Then we find:

#5isqrt(-54) = 5i^2sqrt(54) = -5sqrt(3^2*6) = -5sqrt(3^2)sqrt(6) = -15sqrt(6)#

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