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

2 Answers
Don't Memorise
Apr 7, 2015

#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#

Meave60
Apr 7, 2015

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#

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