How do you rationalize the denominator and simplify #sqrt3/(2+sqrt2)#?

Question

Eva Abramson · Accepted Answer

You multiply by #1#, disguised as #(2-sqrt2)/(2-sqrt2)# #=(sqrt3xx(2-sqrt2))/((2+sqrt2)xx(2-sqrt2))# The denominator is now a special product of the sort: #(A+B)(A-B)=A^2-B^2# #=(sqrt3xx(2-sqrt2))/(2^2-sqrt2^2)=(2sqrt3-sqrt2*sqrt3)/(4-2)# #=(2sqrt3-sqrt6)/2=sqrt3-1/2sqrt6#

Ethan Adkins · Answer

undefined To rationalize the denominator and simplify the expression sqrt(3)/(2+sqrt(2)), we can multiply both the numerator and denominator by the conjugate of the denominator, which is 2-sqrt(2).

By doing this, we get (sqrt(3)/(2+sqrt(2))) * (2-sqrt(2))/(2-sqrt(2)).

Simplifying this expression, we have (2sqrt(3) - sqrt(6))/(4-2sqrt(2)).

To further simplify, we can multiply both the numerator and denominator by the conjugate of the denominator again, which is 4+2sqrt(2).

This gives us ((2sqrt(3) - sqrt(6))/(4-2sqrt(2))) * (4+2sqrt(2))/(4+2sqrt(2)).

Simplifying further, we have (8sqrt(3) + 4sqrt(6) - 4sqrt(6) - 2sqrt(12))/(16 - 8sqrt(2) + 8sqrt(2) - 4*2).

Combining like terms, we get (8sqrt(3) - 2sqrt(12))/(16 - 8).

Simplifying the denominator, we have (8sqrt(3) - 2sqrt(12))/8.

Finally, we can simplify the expression by factoring out a common factor of 2 from the numerator, resulting in 2(4sqrt(3) - sqrt(12))/8.

Simplifying further, we have (4sqrt(3) - sqrt(12))/4.

Therefore, the rationalized and simplified form of sqrt(3)/(2+sqrt(2)) is (4sqrt(3) - sqrt(12))/4.