How do you simplify #(sqrt3-sqrt2)(sqrt15+sqrt12)#?

1 Answer
May 29, 2017

Answer:

See a solution process below:

Explanation:

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

#(color(red)(sqrt(3)) - color(red)(sqrt(2)))(color(blue)(sqrt(15)) + color(blue)(sqrt(12)))# becomes:

#(color(red)(sqrt(3)) xx color(blue)(sqrt(15))) + (color(red)(sqrt(3)) xx color(blue)(sqrt(12))) - (color(red)(sqrt(2)) xx color(blue)(sqrt(15))) - (color(red)(sqrt(2)) xx color(blue)(sqrt(12)))#

Next, use this rule of radicals to multiply the four sets of radicals:

#sqrt(color(red)(a)) * sqrt(color(blue)(b)) = sqrt(color(red)(a) * color(blue)(b))#

#sqrt(3 xx 15) + sqrt(3 xx 12) - sqrt(2 xx 15) - sqrt(2 xx 12)#

#sqrt(45) + sqrt(36) - sqrt(30) - sqrt(24)#

#sqrt(45) + 6 - sqrt(30) - sqrt(24)#

Now, we can rewrite the expression to simplify as:

#sqrt(9 xx 5) + 6 - sqrt(30) - sqrt(4 xx 6)#

#(sqrt(9) xx sqrt(5)) + 6 - sqrt(30) - (sqrt(4) xx sqrt(6))#

#3sqrt(5) + 6 - sqrt(30) - 2sqrt(6)#