How do you simplify #sqrt42 * sqrt12#?

1 Answer
Alan N.
Apr 30, 2016

#6sqrt(14)#

Explanation:

All integers can be expressed as the product of prime numbers. In problems involving finding roots of integers it is often useful to start by expressing the integers as products of prime numbers.

From this example:

#42 = 2*3*7# and #12 = 2*2*3#

Here we are asked to simplify #sqrt(42)*sqrt(12)#

From the above factorisation:

#sqrt(42)*sqrt(12)##=sqrt(2*3*7)*sqrt(2*2*3)#

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

#=sqrt(2*3*7*2*2*3)#

Since #sqrt(a*a) = a# any pair of like factors may be taken out of the #sqrt# sign. Thus the expression becomes:

#= 2*3*sqrt(2*7)#

#=6sqrt(14)#

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