How do you simplify #sqrt18 div sqrt(8 - 3)#?

Question

Eva Abramson · Accepted Answer

#sqrt18/sqrt(8-3)# # = sqrt18/sqrt5# # = sqrt(9*2)/sqrt5# # = (sqrt9*sqrt2)/sqrt5# We used the Identity #sqrt(a*b)=sqrta*sqrtb# # = (3*sqrt2)/sqrt5#           -------------(a) Next, we need to RATIONALISE the denominator.
To do that, we multiply the Numerator and the Denominator by #sqrt5# # = (3*sqrt2)/sqrt5##*##sqrt5/sqrt5# # = (3*sqrt10)/5# The answer can be left in this form, but if you want to find the numerical value of the expression, we can substitute the approximate values of #sqrt2# and #sqrt5# in (a) We get #((3) * (1.414))/2.236 ~ 1.9#

Eva Adamson · Answer

The answer is #3sqrt(2/5)# or #(3sqrt10)/5#. #sqrt18##-:##sqrt(8-3)# = #sqrt18/sqrt(8-3)# = #(sqrt2sqrt9)/(sqrt5)# = (#sqrt9=3#) #(3sqrt2)/sqrt5# =  #3sqrt(2/5)# To remove #sqrt5# from the denominator, multiply the numerator and denominator by #sqrt5#. #(3sqrt2)/(sqrt5)*sqrt5/sqrt5# = #(3sqrt2*sqrt5)/5# = #(3sqrt10)/5#

Ethan Adkins · Answer

undefined To simplify sqrt18 div sqrt(8 - 3), we can simplify each square root separately and then divide the results.

First, let's simplify sqrt18. We can rewrite 18 as 9 * 2, and since 9 is a perfect square, we can take its square root. The square root of 9 is 3, so sqrt18 simplifies to 3sqrt2.

Next, let's simplify sqrt(8 - 3). We subtract 3 from 8, which gives us 5. The square root of 5 cannot be simplified further, so sqrt(8 - 3) remains as sqrt5.

Now, we can divide the simplified square roots. 3sqrt2 div sqrt5 can be written as (3/1) * (sqrt2/sqrt5). To simplify this further, we rationalize the denominator by multiplying both the numerator and denominator by sqrt5. This gives us (3 * sqrt2 * sqrt5) / (1 * sqrt5 * sqrt5).

Simplifying the expression, we get (3 * sqrt(2 * 5)) / (1 * 5), which simplifies to (3 * sqrt10) / 5.

Therefore, sqrt18 div sqrt(8 - 3) simplifies to (3 * sqrt10) / 5.