How do you simplify #sqrt14/(3sqrt35)#?

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Answer 1 · 2016-09-06T15:49:30

#sqrt10/15#

Write as #" "sqrt(2xx7)/(3sqrt(5xx7))#

#(sqrt(2)xx cancel(sqrt(7)))/(3xxsqrt(5)xx cancel(sqrt(7)))#

#sqrt(2)/(3sqrt(5))#

If at all possible mathematicians like to 'get rid' of any roots in the denominator (bottom number).

Multiply by 1 but in the form #1=sqrt5/sqrt5#

#sqrt(2)/(3sqrt(5))xx1" " ->" "sqrt(2)/(3sqrt(5))xxsqrt5/sqrt5#

but #" "sqrt5xxsqrt5 =5#

#(sqrt2sqrt5)/(3xx5)#

#sqrt10/15#

Answer 2 · 2016-09-06T15:51:23

#(sqrt(10)) / (15)#

We have: #(sqrt(14)) / (3 sqrt(35))#

Let's express the radicals as products:

#= (sqrt(2 cdot 7)) / (3 cdot sqrt(5 cdot 7))#

#= (sqrt(2) cdot sqrt(7)) / (3 cdot sqrt(5) cdot sqrt(7))#

We can cancel the #sqrt(7)# terms:

#= (sqrt(2)) / (3 sqrt(5))#

Finally, let's rationalise the denominator:

#= (sqrt(2)) / (3 sqrt(5)) cdot (sqrt(5)) / (sqrt(5))#

#= (sqrt(10)) / (3 cdot 5)#

#= (sqrt(10)) / (15)#