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Find the consecutive numbers whose squares have the sum 85.
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Let the two consecutive natural numbers be ‘x’ and ‘x + 1’
⇒ Given that the sum of their squares is 85.
Then by hypothesis, we get
ЁЭСе2 + (ЁЭСе + 1)2 = 85
⇒ ЁЭСе2 + ЁЭСе2 + 2ЁЭСе + 1 = 85
⇒ 2ЁЭСе2 + 2ЁЭСе + 1 − 85 = 0
⇒ 2ЁЭСе2 + 2ЁЭСе + 84 = 0 ⇒ 2 [ЁЭСе2 + ЁЭСе − 42 ] = 0
⇒ ЁЭСе2 + 7ЁЭСе - 6ЁЭСе - 42 = 0 [by the method of factorisation]
⇒ ЁЭСе(ЁЭСе + 7) - 6(ЁЭСе + 7) = 0
⇒ (ЁЭСе - 6)(ЁЭСе + 7) = 0
⇒ ЁЭСе = 6 or ЁЭСе = 7
Case i: if x = 6 x + 1 = 6 + 1 = 7
Case ii: If x = 7 x + 1 = -7 + 1 = -6
∴ The consecutive numbers that the sum of this squares be 85 are 6,7 and -6, -7.
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