How do you evaluate #\frac { 7} { 2+ 2\sqrt { 3} }#?

Question

Abigail Allen · Accepted Answer

#7/(2+2sqrt3)=(7sqrt3-7)/4# #7/(2+2sqrt3)=7/(2sqrt3+2)# = #7/(2sqrt3+2)xx(2sqrt3-2)/(2sqrt3-2)# and using #(a+b)(a-b)=(a^2-b^2)#, we get #(7(2sqrt3-2))/((2sqrt3)^2-2^2)# = #(14sqrt3-14)/(12-4)# = #(14sqrt3-14)/8# and dividing numerator and denominator by #2# = #(7sqrt3-7)/4#

Ethan Adkins · Answer

undefined To evaluate the expression $\frac{7}{2+2\sqrt{3}}$, we can rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is $2-2\sqrt{3}$. Simplifying this expression gives us $\frac{7(2-2\sqrt{3})}{(2+2\sqrt{3})(2-2\sqrt{3})}$. Expanding the denominator further, we have $\frac{7(2-2\sqrt{3})}{4-12}$. Simplifying the denominator gives us $\frac{7(2-2\sqrt{3})}{-8}$, which can be further simplified to $\frac{-7(2\sqrt{3}-2)}{8}$. Finally, simplifying the expression gives us $\frac{-7\sqrt{3}+7}{8}$.