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Find two consecutive positive integers , sum of whose squares is 365 |
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Answer» let the no. be x and x+1 then we have x²+(x+1)²=365 2x²+2x-364=0 x²+x-182=0 x²+14x-13x-182=0 (x+14)(x-13)=0 That gives x=13 Avoiding negative value Hence, one number is 13 and other one 14. Let two consecutive positive integer be x and x + 1.According to questionx2 + (x + 1)2 = 365⇒ x2 + x2 + 2x + 1 = 365⇒ 2x2 + 2x - 364 = 0⇒ x2 + x - 182 = 0Let us split middle termx2 + 14x - 13x - 182 = 0⇒ (x + 14) (x - 13) = 0x = -14 or x = 13 [ - ve root being rejected ]Hence required numbers are 13 and 14 13 and 14.... |
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