How do you simplify #(6sqrt5)/(5sqrt3)#?

1 Answer

Answer:

#((2sqrt(15))/(5))#

Explanation:

Let's first talk about what it is about this fraction that needs simplification. When we divide by an integer (so a number such as 2 or -4), we are dividing by something that is definite and certain. When we divide by an irrational number (such as a square root), we are now dividing by something that as neat and tidy as it was by dividing by an integer.

So by "simplify", the question is really asking you to get rid of the #sqrt(3)# in the denominator.

Simplification questions often use creative forms of the number 1. Remember that 1 = anything/that same anything. So using the appropriate "anything" is the key.

The #sqrt(3)# in the denominator will decide what the appropriate "anything" should be.

Remember that #sqrt(x)# times #sqrt(x)# = x. So we can multiply #sqrt(3)# by itself and get rid of the square root.

But we can't simply multiply the denominator by #sqrt(3)# because that would change the value of the fraction we are working on. We need to multiply the whole fraction by 1. So multiplying #(6sqrt(5))/(5sqrt(3))# by #(sqrt(3)/sqrt(3))# (remember that #(sqrt(3)/sqrt(3))#=1) will give us

#((6sqrt(5)sqrt(3))/(5sqrt(3)sqrt(3)))#

which equals

#((6sqrt(15))/(5*3))#

Dividing the denominator 3 into the numerator 6 gives

#((2sqrt(15))/(5))#