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Question 11
Find the values of a and b in each of the following
(i)5+237+43=a63
(ii)353+25=a51911
(iii)2+33223=2b6
(iv)7+575757+5=a+7115b

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Solution

(i) We have 5+237+43=a63
For rationalizing the above equation, we multiply numerator and denominator of LHS by 7 - 4 3, we get,
5+237+43×743743=a635(743)+23(743)72(43)2=a63
[Using identity,(a+b)(ab)=a2b2]
35203+143244948=a631163=a63a=11

(ii) We have, 353+25=a51911
For rationalizing the above equation, we multiply numerator and denominator of LHS by 325, we get
(35)3+25×325325=a519113(325)5(325)(3)225)2=a51911
[Using identity, (ab)(a+b)=a2b2]
96535+1094×5=a519111995920=a51911199511=a5191195111911=a519119511=a5a=911

(iii) We have, 2+33223=2b6
For rationalising the above equation, we multiply numerator and denominator of LHS by 32+23, we get
2+33223×32+2332+23=2b62(32+23)+3(32+23)(32)2(23)2=2b6
[Using identity,(ab)(a+b)=a2b2]
6+26+36+61812=2b612+566=2b62+566=2b6b6=566b=56

(iv) We have, 7+575757+5=a+7115b
(7+5)2(75)2(75)(7+5)=a+7115b[72+(5)2+2×7×5](72+(5)22×7×5]72(5)2=a+7115b
Using identity,(a+b)2=a2+2ab+b2(ab)2=a22ab+b2and(ab)(a+b)=a2b2
49+5+145495+145495=a+7115b28544=a+7115b0+7115=a+7115b
On comparing both sides, we get
a = 0 and b = 1


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