Rationalize the denominator.?

Question

1/(1+rad3-rad5)

Answer 1

1 / ( 1 + √3 - √5 ) =
( 1 - √3 + √5 ) / ( 1 + √3 - √5 ) ( 1 - √3 + √5 ) =
( 1 - √3 + √5 ) / ( ( 1 - √3 + √5 ) + √3( 1 - √3 + √5 ) - √5( 1 - √3 + √5 ) ) =
( 1 - √3 + √5 ) / ( 1 - √3 + √5 + √3 - √3√3 + √3√5 - √5 + √3√5 - √5√5 ) =
- ( 1 - √3 + √5 ) / (7 - 2√15 ) =
- ( 1 - √3 + √5 ) (7 + 2√15 ) / (7 - 2√15 ) (7 + 2√15 ) =
- ( (7 + 2√15 ) - √3(7 + 2√15 ) + √5(7 + 2√15 ) ) / (7(7 + 2√15 ) - 2√15(7 + 2√15 ) ) =
- ( (7 + 2√15 - 7√3 - 2√3√15 + 7√5 + 2√5√15 ) ) / ((49 - 4*15 ) =
- ( 7 + 2√15 - 7√3 - 2*3√5 + 7√5 + 2*5√3 ) / ((49 - 4*15 ) =
( 7 + 3√3 + √5 + 2√15 ) / 11 =

Answer 2

To rationalize the denominator, you must multiply numerator and denominator by its conjugate to remove the radical. In this particular instance, you have two radicals and must do it twice. So pick your radical to remove first and multiply by the suitable conjugate. In this case, I choose to remove the sqrt(5) radical first, so I multiply both top and bottom to get

( 1 / (1 + sqrt(3) - sqrt(5)) ) * ( (1 + sqrt(3) + sqrt(5)) / (1 + sqrt(3) + sqrt(5)) )
= (1 + sqrt(3) + sqrt(5)) / ( (1 + sqrt(3))^2 - 5)
= (1 + sqrt(3) + sqrt(5)) / (1 + 2*sqrt(3) + 3) - 5)
= (1 + sqrt(3) + sqrt(5)) / (2*sqrt(3) - 1)

After the above simplification, we may now get rid of the final radical by multiplying by the conjugate of the simplified denominator to get

(1 + sqrt(3) + sqrt(5)) * (2*sqrt(3) + 1) / ((2*sqrt(3) - 1)(2*sqrt(3) + 1))
= (2*sqrt(3) + 1 + 2*3 + sqrt(3) + 2*sqrt(15) + sqrt(5)) / ((2*sqrt(3))^2 - 1^2)
= (7 + 3*sqrt(3) + sqrt(5) + 2*sqrt(15)) / (12 - 1)
= (7 + 3*sqrt(3) + sqrt(5) + 2*sqrt(15)) / 11

Answer 3

1/(1-rad2)