Compare the surds
(i) `A=sqrt(10)-sqrt(5),B=sqrt(19)-sqrt(14)`
(ii) `P=sqrt(10)+sqrt(5),Q=sqrt(8)+sqrt(7)`

Answer

N/a

Explanation

(i) `A=sqrt(10)-sqrt(5)=((sqrt(10)-sqrt(5))(sqrt(10)+sqrt(5)))/(sqrt(10)+sqrt(5))=(10-5)/(sqrt(10)+sqrt(5))=(5)/(sqrt(10)+sqrt(5))` . . .(1)
`B=sqrt(19)-sqrt(14)=((sqrt(19)-sqrt(14))(sqrt(19)+sqrt(14)))/(sqrt(19)+sqrt(14))=(19-14)/(sqrt(19)+14)=(5)/(sqrt(19+sqrt(14))` . . . (2)
The numerator of each A and B is same i.e., 5. But denominator
`sqrt(19)+sqrt(14)gtsqrt(10)+sqrt(5)`
`:.(1)/(sqrt(19)+sqrt(14))lt(1)/(sqrt(10)+sqrt(5))`
`rArr(5)/(sqrt(19)+sqrt(14))lt(5)/(sqrt(10)+sqrt(5))`
`rArr(5)/(sqrt(19)-sqrt(14))lt(5)/(sqrt(10)-sqrt(5))` [from (1) and (2)]
(ii) Since there is a positive sign in
`P=sqrt(10)+sqrt(5)andQ=sqrt(8)+sqrt(7)`
`:."Squaring P and Q, we get"`
`P^(2)=105+2sqrt(50)=15+2sqrt(50)`
`Q^(2)=8+5+2sqrt(56)=15+2sqrt(56)`
As `sqrt(56)gtsqrt(50)" "rArr" "2sqrt(56)gt2sqrt(50)" "rArr" "15+2sqrt(56)gt15+2sqrt(50)`
`rArrQ^(2)gtP^(2)" "rArr" "QgtP" "rArr" "sqrt(8)+sqrt(7)gtsqrt(10)+sqrt(5)`