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Find the squares of the following numbers using the identity (a-b)2= a2-2ab+b2(i) 395(ii) 995(iii)495(iv) 498 |
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Answer» (i) 395 We know, (a-b)2= a2-2ab+b2 395 = (400-5)2 = (400)2 + 52 – 2 (400) (5) = 160000 + 25 – 4000 = 156025 (ii) 995 We know, (a-b)2= a2-2ab+b2 995 = (1000-5)2 = (1000)2 + 52 – 2 (1000) (5) = 1000000 + 25 – 10000 = 990025 (iii) 495 We know, (a-b)2= a2-2ab+b2 495 = (500-5)2 = (500)2 + 52 – 2 (500) (5) = 250000 + 25 – 5000 = 245025 (iv) 498 We know, (a-b)2= a2-2ab+b2 498 = (500-2)2 = (500)2 + 22 – 2 (500) (2) = 250000 + 4 – 2000 = 248004 |
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