How do you rationalize the denominator and simplify #sqrt(3/2)#?

3 Answers
May 5, 2018

Answer:

I got as far as this:

Explanation:

Let us write it as:

#sqrt(3)/sqrt(2)#

multiply and divide by #sqrt(2)#

#sqrt(3)/sqrt(2)color(red)(sqrt(2)/sqrt(2))=(sqrt(3)sqrt(2))/2==sqrt(3*2)/2=sqrt(6)/2#

May 5, 2018

Answer:

#sqrt6/2#

Explanation:

#sqrt (3/2)#

#:.=sqrt 3/sqrt 2#

#:.=sqrt 3/sqrt 2xx sqrt 2/sqrt 2#

#:.=sqrt2xxsqrt2=2#

#:.=(sqrt(2xx3))/2#

#:.=sqrt6/2#

May 5, 2018

Answer:

#sqrt6/2=1/2sqrt6#

Explanation:

#"using the "color(blue)"laws of radicals"#

#•color(white)(x)sqrt(a/b)hArrsqrta/sqrtb#

#•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)#

#•color(white)(x)sqrtaxxsqrta=a#

#sqrt(3/2)=sqrt3/sqrt2#

#"to eliminate the radical on the denominator multiply"#
#"the numerator/denominator by "sqrt2#

#rArrsqrt3/sqrt2xxsqrt2/sqrt2=(sqrt3xxsqrt2)/(sqrt2xxsqrt2)=sqrt6/2=1/2sqrt6#