How do you rationalize the denominator and simplify #3/sqrt12#?

2 Answers
Mar 20, 2016

Answer:

# = sqrt12/4#

Explanation:

#3 / sqrt12#

#(3 * color(blue)(sqrt12))/(( sqrt12 * color(blue)( sqrt12))#

# = ((3sqrt12))/12#

# = (cancel3sqrt12)/cancel12#

# = sqrt12/4#

Mar 20, 2016

Answer:

Bit more explanation

#= (sqrt(12))/4#

Explanation:

If you multiply a value by 1 you do not change its intrinsic value.

The thing is, 1 can come in many forms. For example#" " 3/3#.

So you can write 1 as #1=sqrt(12)/sqrt(12)#

Given:#" "3/sqrt(12)#

Multiply by 1 but in the form #1=sqrt(12)/sqrt(12)#

#" "3/sqrt(12)xxsqrt(12)/(sqrt(12))#

#=(3xxsqrt(12))/(sqrt(12)xxsqrt(12))#

#=(3sqrt(12))/12 #

Divide top and bottom by 3

#=(3sqrt(12)-:3)/(12-:3) #

#=(sqrt(12))/4#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("If you really wanted to be absolutely correct about the answer ")#
#color(red)("consider this:")#

#sqrt(12)# is a number that when multiplied by itself results in +12

However# sqrt(12)# could be negative. Think about:

# (-3)xx(-3) = +9 = (+3)xx(+3)#

So should the answer be:

#color(green)(+-sqrt(12)/4)#

#color(red)("No!")#

The question gives: # (sqrt(12))/4#

The absence of any sign means that it is positive. Consequently the questions single #sqrt(12)# is positive. So the solutions single #sqrt(12)# has to be positive as well.