How do you simplify #sqrt8*sqrt10#?

3 Answers
Nov 21, 2017

Answer:

Arrange the equation and get #4sqrt5#

Explanation:

#sqrt8timessqrt10 = sqrt(80)# under one square root. Further,

#=sqrt(16times5) = 4sqrt(5)#

This is the shortest form. Your answer is #4sqrt5#

Nov 21, 2017

Answer:

#sqrt8sqrt10=color(blue)(4sqrt5#

Explanation:

Simplify:

#sqrt8sqrt10#

Prime factorize #8#.

#sqrt((2xx2)xx2)*sqrt10#

#2sqrt2*sqrt10#

#2sqrt(2xx10)#

#2sqrt(20)#

Prime factorize #20#.

#2sqrt((2xx2)xx5)#

Simplify.

#2xx2sqrt5#

Simplify.

#4sqrt5#

Nov 21, 2017

Answer:

#4sqrt(5)#

Explanation:

short answer:
you know that when multiplying roots they can go 'in each other' thus:
#sqrt(8)* sqrt (10)# = #sqrt(8*10)#

then break down each number to it's primary numbers

#sqrt((2*2*2)*(2*5))#

notice that we have numbers that are repeated twice (because here we have a square root)
#sqrt((2*2)*(2*2)*5)#

then take them out of the square root. I took them out one at a time.
#2sqrt((2*2)*5)#
Note that when you take them out you put only one repeated term

take the second 2 out and remember this is all multiplication so 2 is multiplied by the 2 in front of the square root.

#(2*2)sqrt(5)#

the #sqrt(5)# cannot be simplified, so we leave it as it is. (there is not a whole numbers that we can multiply by itself to get 5 )

so we are left with just simplifying the front #2*2 = 4# thus

#4sqrt(5)#

Note: this could be also solved as:
#sqrt((2*2)*(2*2)*5) = sqrt(4*4*5)#
thus 4 will be our repeated twice number, then we simply would take it out to get :
#4sqrt(5)#

The second way would be more useful when dealing with larger numbers.

I hope this helps. thank you.