How do you simplify #\frac { - 4+ 4\sqrt { 3} } { 2+ \sqrt { 2} }#?

1 Answer
smendyka
Aug 13, 2017

See a solution process below

Explanation:

To start the simplification process we need to rationalize the denominator. Or, in other words, remove the radicals from the denominator.

We can use the rule #(color(red)(x) + color(blue)(y))(color(red)(x) - color(blue)(y)) = color(red)(x)^2 - color(blue)(y)^2# to find the appropriate form of #1# to multiply the fraction by:

#(color(red)(2) - color(blue)(2sqrt(2)))/(color(red)(2) - color(blue)(2sqrt(2))) xx (-4 + 4sqrt(3))/(color(red)(2) + color(blue)(2sqrt(2))) =>#

#((color(red)(2) xx -4) + (color(red)(2) xx 4sqrt(3)) + (color(blue)(2sqrt(2)) xx 4) - (color(blue)(2sqrt(2)) xx 4sqrt(3)))/(color(red)(2)^2 - (color(blue)(2sqrt(2)))^2) =>#

#(-8 + 8sqrt(3) + 8sqrt(2) - 8sqrt(6))/(4 - (4 * 2)) =>#

#(8(-1 + sqrt(3) + sqrt(2) - sqrt(6)))/(4 - 8) =>#

#(8(-1 + sqrt(3) + sqrt(2) - sqrt(6)))/(-4) =>#

#(color(red)(cancel(color(black)(8)))-2(-1 + sqrt(3) + sqrt(2) - sqrt(6)))/color(red)(cancel(color(black)(-4))) =>#

#-2(-1 + sqrt(3) + sqrt(2) - sqrt(6))#

Or

#2(1 - sqrt(3) - sqrt(2) + sqrt(6))#

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