How do you simplify #sqrt42 * sqrt12#?

Question

How do you simplify #sqrt42  * sqrt12#?

Julian Avila · Accepted Answer

#6sqrt(14)#

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)#

Gianna Caruso · Answer

undefined To simplify $ \sqrt{42} \times \sqrt{12} $, you multiply the numbers inside the square roots together:

$$ \sqrt{42} \times \sqrt{12} = \sqrt{42 \times 12} $$

Then, you simplify the expression inside the square root:

$$ 42 \times 12 = 504 $$

So,

$$ \sqrt{42} \times \sqrt{12} = \sqrt{504} $$

Next, you find the largest perfect square that divides evenly into 504, which is 36:

$$ 504 = 36 \times 14 $$

Therefore,

$$ \sqrt{504} = \sqrt{36 \times 14} $$

$$ = \sqrt{36} \times \sqrt{14} $$

$$ = 6 \times \sqrt{14} $$

So,

$$ \sqrt{42} \times \sqrt{12} = 6 \sqrt{14} $$