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Question

Using the formula for squaring a binomial, evaluate the following:

(i) (102)2        (ii) (99)2        (iii) (1001)2        (iv) (999)2        (v) (703)2


Solution

(i) (102)2=(100+2)2=(100)2+2×100×2+(2)2{(a+b)2=a2+2ab+b2}=10000+400+4=10404(ii) (99)2=(1001)2=(100)22×100×1+(1)2{(ab)2=a22ab+b2}=10000200+1=10001200=9801(iii) (1001)2=(1000+1)2{(a+b)2=a2+2ab+b2}=(1000)2+2×1000×1+(1)2=1000000+2000+1=1002001(iv) (999)2=(10001)2{(ab)2=a22ab+b2}=(1000)22×1000×1+(1)2=10000002000+1=10000012000=998001(v) (703)2=(700+3)2=(700)2+2×700×3+(3)2{(a+b)2=a2+2ab+b2}=490000+4200+9=494209

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