How do you simplify #sqrt10sqrt8#?

2 Answers
Don't Memorise
Jun 15, 2015

# = color(red)(4sqrt5#

Explanation:

#sqrt10sqrt8 = sqrt(80)#

# sqrt 80 = sqrt(4 . 4 . 5 )#

# = color(red)(4sqrt5#

Bill K.
Jun 17, 2015

If you ever wondered why #sqrt(a)sqrt(b)=sqrt(ab)# (for #a,b \geq 0#), the reason follows from the commutative and associative properties of multiplication: #(sqrt(a)sqrt(b))^2=(sqrt(a)sqrt(b))*(sqrt(a)sqrt(b))#

#=(sqrt(a)sqrt(a))*(sqrt(b)sqrt(b))=a*b#.

But #sqrt(ab)# is the unique nonnegative number whose square is #ab#. Hence #sqrt(a)sqrt(b)=sqrt(ab)#.

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
1247 views around the world
You can reuse this answer
Creative Commons License