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

1 Answer
Aug 1, 2017

See a solution process below:

Explanation:

We can rationalize the denominator by using this rule for multiply quadratics:

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

Substitute: #color(red)(7)# for #color(red)(x)# and #color(blue)(sqrt(5))# for #color(blue)(y)#

#(color(red)(7) + color(blue)(sqrt(5)))/(color(red)(7) + color(blue)(sqrt(5))) xx (color(red)(7) + color(blue)(sqrt(5)))/(color(red)(7) - color(blue)(sqrt(5))) =>#

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

#(49 + 14sqrt(5) + 5)/(49 - 5) =>#

#(54 + 14sqrt(5))/44 =>#

#(2(27 + 7sqrt(5)))/(2 xx 22) =>#

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