How do you rationalize the denominator of #(7-sqrt5)/(7+sqrt5)#?

1 Answer
Lotusbluete
Jan 22, 2016

#(27- 7sqrt(5))/22#

Explanation:

To rationalize the denominator, you should take advantage of the formula

#(a+b)(a-b) = a^2 - b^2#

To do so, you need to expand your fraction with #7 - sqrt(5)#:

#(7 - sqrt(5))/(7 + sqrt(5)) = ((7 - sqrt(5))color(blue)((7 - sqrt(5))))/((7 + sqrt(5))color(blue)((7 - sqrt(5)))) = (7 - sqrt(5))^2/((7 + sqrt(5))(7 - sqrt(5)))#

To simplify the numerator, apply the formula

#(a-b)^2 = a^2 - 2ab + b^2#

To simplify the denominator, apply the formula

#(a+b)(a-b) = a^2 - b^2#

Thus, you will get:

#... = (7^2 - 2*7*sqrt(5) + (sqrt(5))^2)/ (7^2 - (sqrt(5))^2) = (49 - 14 sqrt(5) + 5)/(49-5) #

# = (54- 14 sqrt(5))/44 = (2(27-7sqrt(5)))/(2*22) = (cancel(2)(27-7sqrt(5)))/(cancel(2)*22) #

# = (27- 7sqrt(5))/22#

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