How do you simplify #[sqrt 18] - 2[sqrt 2]#?

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Answer 1 · 2016-05-09T20:55:08

#sqrt18-2sqrt2=sqrt2#

We will make use of the fact that #sqrt(ab)=sqrta*sqrtb# when #a# and #b# are positive.

From this, we can see that #sqrt18=sqrt(9*2)=sqrt9*sqrt2=3*sqrt2=3sqrt2#.

Thus, #sqrt18-2sqrt2=3sqrt2-2sqrt2=sqrt2(3-2)=sqrt2(1)=sqrt2#.

Answer 2 · 2016-05-09T21:07:54

#sqrt 18-2sqrt 2=color(blue)sqrt 2#

#sqrt 18-2sqrt 2#

Simplify #sqrt 18# by writing its prime factors.

#sqrt(2xx3xx3)-2sqrt 2#

#sqrt(2xx3^2)-2sqrt 2#

Apply the square root rule #sqrt(a^2)=a#.

#3sqrt 2-2sqrt 2#

Simplify.

#sqrt 2#