How do you simplify #sqrt(-50)#?

1 Answer
Jul 5, 2015

Answer:

I found: #sqrt(-50)=5isqrt(2)#

Explanation:

Here you have a problem...
in fact you cannot solve a square root with a negative argument or, better, you cannot find a real number which is solution of your square root.

What you can do it is try to find a solution somewhere else...in the place of imaginary numbers!
First of all you manipulate your root as:
#sqrt(-50)=sqrt(-1*2*25)=sqrt(-1)sqrt(25)sqrt(2)=5sqrt(-1)sqrt(2)#
you now introduce a new entity: the imaginary unit #i# given as:
#i=sqrt(-1)#
so you can write:
#sqrt(-50)=5isqrt(2)# which is:

1] solution of your problem; in fact if you square it you get:
#(5isqrt(2))^2=25*-1*2=-50# that is your original radicand!

2] it is an imaginary number (it contains #i#).