How do you simplify #3sqrt50*sqrt22#?

3 Answers
May 5, 2018

Answer:

#30sqrt11#

Explanation:

#3sqrt50*sqrt22#

#3sqrt(25*2)*sqrt22 rarr 25# is a perfect square and can be taken out of the radical

#3*5sqrt2*sqrt22#

#15sqrt2*sqrt22#

#15sqrt(2*22)#

#15sqrt44#

#15sqrt(4*11) rarr 4# is a perfect square

#15*2sqrt11#

#30sqrt11#

Answer:

I'm going to try to write out your question mathematically:
#root(3)(50xxsqrt22)#

Explanation:

To solve this exceedingly complicated question, let's start by converting everything into indices.

But first, we should simplify your equation:
#root(3)(sqrt55000)#
Now the question will be written in indicial form as:
#(55000^(1/2))^(1/3)#

By the power law of indices #-># #(a^m)^n=a^(mn)#, this would make the indices #1/2xx1/3=1/6#.
Therefore, the question will now become:
#55000^(1/6)#
#=root(6)55000#
#=50root(6)22#
In surd form.

I hope this helps!

May 5, 2018

Answer:

#30sqrt11#

Explanation:

#3sqrt50*sqrt 22#

#:.=3 sqrt(2*5*5)*sqrt(2*11)#

#:.=sqrt 5*sqrt 5=5#

#:.=3*5 sqrt 2*sqrt 2*sqrt 11#

#:.=15 sqrt 2*sqrt 2*sqrt 11#

#:.=sqrt2*sqrt2=2#

#:.=2*15*sqrt 11#

#:.=30 sqrt11#