How do you simplify #sqrt 8 /( 2 sqrt3)#?
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#(sqrt8)/(2sqrt 3)=color(blue)((sqrt 6)/3)#
#(sqrt 8)/(2sqrt 3)#
Simplify #sqrt 8#.
#sqrt 8=sqrt(2xx2xx2)=sqrt(2^2xx 2)=2sqrt2#
Rewrite the fraction.
#(2sqrt2)/(2sqrt 3)#
Rationalize the denominator by multiplying the numerator and denominator by #sqrt 3#.
#(2sqrt2)/(2sqrt 3)xx(sqrt3)/(sqrt 3)#
Simplify.
#(2sqrt2sqrt3)/(2xx3)#
Simplify.
#(2sqrt6)/(2xx3)#
Simplify.
#(cancel2sqrt6)/(cancel2xx3)#
Simplify.
#(sqrt 6)/3#
#8=2^3#
#sqrt (8) = 2^(3/2)#
Therefore we have
#(2^(3/2).2^(-1))/sqrt (3)#
Add the exponent coefficients for 2
#(2^(1/2))/sqrt (3)#
Same as #sqrt(2/3)#
#sqrt8/(2sqrt3)#
We could see that
#sqrt8=sqrt(4*2)#
So
#=sqrt(4*2)/(2sqrt3_#
#=(cancel2sqrt2)/(cancel2sqrt3)#
#=sqrt2/sqrt3=sqrt(2/3)#
But wait ! We could not have irrational numbers in the denominator.
So,rationalize the denominator by multiplying with #sqrt3/sqrt3#
#sqrt2/sqrt3*sqrt3/sqrt3#
#=sqrt6/3#
