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Sa Exercise 1.2 am Ql. Rationalize the denominator of following: 13 (i) TB Solution: 13 13. 4-v3 pe ae HB A 13(4-V3) (4+43)(4-3) _ 13(4-V3) ~ tate (-:(a+b)(a—b)=a" -b*) _ 134-13) 16-3 _ ¥5(4—-13) % 13 2 42.48 Ans. 4+13 Ea j : (ii) ae Solution: V5x2—V5x1 S Fis ABB py, 6-42 ©) ree Solution: 6r42 644N2 6-442 ___(6-4V2 ; | (6+ H2)6-412) Using identities (ab)? =a? +6? ~2ab and (a+b}(a—b)=a? ~B, the above eq. becomes 6-4v2 (6) + (42)? ~216)(4 V2: am (FH = i | Rationalizing. the den 36+(16x2)-48y2 ~ ~"36~(16%2) 36+32-48V2 ~ 36-32 68-482 ee _ #«(17-12V2) é # = 17-122 Ans,” ¥3-¥2 B+2 Solution: Rationalizing the denominator by multiplying ang dividing the fraction with conjugate of denominator, ~ 8-2 _ ¥3-v2, V3-v2 Vaei2 | WB+V2 Va—V2 _ __ WJ? Wa +W2)3—V2) Using identities (a—b)° =a? +5*—2ab and (a+b)(a—b)=a?—b?,, the above eq. becomes Boi _ AEP Bie Way (2 i (vy) = 342-2VBx2 3-2 _ 5-26 Ya = 5-2N6 Ans. 43 Fae Solution; lominator by multiplying. both ‘ominator by the conjugate of Wi 4 eds B Fae ao ~ HaW7 V5: = Naw —W5) 7-5 2 ANB7 ~ 5) numerator and dent denominator, 
