How do you simplify #sqrt3/(sqrt6 -1) - sqrt3/(sqrt6 + 1)#?

1 Answer
Lotusbluete
Feb 8, 2016

#(2 sqrt(3)) / 5#

Explanation:

You need to find the least common denominator to be able to subtract the two fractions.

In your case, the least common denominator is #(sqrt(6) - 1)(sqrt(6) + 1)#, so you need to expand the first fraction with #(sqrt(6) + 1)# and the second one with #(sqrt(6) - 1)#.

#sqrt(3) / (sqrt(6) - 1) - sqrt(3) / (sqrt(6) + 1) = (sqrt(3)color(blue)((sqrt(6) + 1))) / ((sqrt(6) - 1)color(blue)((sqrt(6) + 1))) - (sqrt(3)color(green)((sqrt(6) - 1))) / ((sqrt(6) + 1)color(green)((sqrt(6) - 1))) #

# = (sqrt(3)(sqrt(6) + 1) - sqrt(3)(sqrt(6) - 1)) / ((sqrt(6) - 1)(sqrt(6) + 1))#

... use the formula #(a + b)(a - b) = a^2 - b^2# to simplify the denominator...

# = (sqrt(3) * sqrt(6) + sqrt(3) - sqrt(3) * sqrt(6) + sqrt(3)) / ((sqrt(6))^2 - 1^2)#

# = (cancel(sqrt(3) * sqrt(6)) + sqrt(3) - cancel(sqrt(3) * sqrt(6)) + sqrt(3)) / (6 - 1)#

# = (2 sqrt(3)) / 5#

# = 2/5 sqrt(3)#

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