How do you rationalize the denominator and simplify #(2sqrt12 - sqrt5) /( sqrt5 + 4sqrt3)#?

1 Answer
Aug 12, 2017

Answer:

See a solution process below:

Explanation:

To rationalize the fraction we need to use the rule:

#(color(red)(x) + color(blue)(y))(color(red)(x) - color(blue)(y)) = color(red)(x)^2 - color(blue)(y)^2#

Because the denominator is the form: #(color(red)(sqrt(5)) + color(blue)(4sqrt(3)))#

We need to use: #(color(red)(sqrt(5)) - color(blue)(4sqrt(3)))# as the numerator and denominator for the appropriate form of #1:

#(2sqrt(12) - sqrt(5))/((color(red)(sqrt(5)) + color(blue)(4sqrt(3)))) xx ((color(red)(sqrt(5)) - color(blue)(4sqrt(3))))/((color(red)(sqrt(5)) - color(blue)(4sqrt(3)))) =>#

#(2sqrt(12)sqrt(5) - (4 * 2)sqrt(12)sqrt(3) - (sqrt(5))^2 + 4sqrt(5)sqrt(3))/((color(red)(sqrt(5)))^2 - (color(blue)(4sqrt(3)))^2) =>#

#(2sqrt(60) - 8sqrt(36) - 5 + 4sqrt(15))/(5 - (16 * 3)) =>#

#(2sqrt(4 * 15) - (8 * 6) - 5 + 4sqrt(15))/(5 - 48) =>#

#(2sqrt(4)sqrt(15) - 48 - 5 + 4sqrt(15))/(-43) =>#

#((2 * 2)sqrt(15) - 53 + 4sqrt(15))/(-43) =>#

#(4sqrt(15) - 53 + 4sqrt(15))/(-43) =>#

#(-53 + 4sqrt(15) + 4sqrt(15))/(-43) =>#

#(-53 + 8sqrt(15))/(-43) =>#

#(-53)/(-43) + (8sqrt(15))/(-43) =>#

#53/43 - (8sqrt(15))/(43)#