How do you simplify #sqrt50/(3sqrt2)#?

3 Answers
Apr 15, 2018

Answer:

#5/3#

Explanation:

#sqrt50/(3sqrt2)=#

#sqrt(25*2)/(3sqrt2)=#

#(5cancelsqrt2)/(3cancelsqrt2)=#

#5/3#

Apr 15, 2018

Answer:

#5/3#

Explanation:

#"using the "color(blue)"law of radicals"#

#•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)#

#rArrsqrt50=sqrt(25xx2)=sqrt25xxsqrt2=5sqrt2#

#rArr(sqrt50)/(3sqrt2)=(5cancel(sqrt2))/(3cancel(sqrt2))=5/3#

Apr 15, 2018

Answer:

#+-5/3#

Explanation:

Let's start by thinking about the number #50#, not #sqrt50#. We know its factors are:

  • #25# and #2#
  • #10# and #5#

So the factors of #sqrt50# are:

  • #color(blue)(sqrt25)# and #color(blue)(sqrt2)#
  • #sqrt10# and #sqrt5#

We can use the first set of factors to rewrite our expression as

#color(blue)(sqrt25*cancelsqrt2)/(3cancelsqrt2)#

The #sqrt2# will cancel on the top and bottom, and we're left with

#sqrt25/3#

Which can be simplified to

#+-5/3#

Hope this helps!