Question

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Find Two Consecutive Natural Numbers Such That The Sum Of Their Squares Is 61.....?​

Answer 1

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Let us consider first natural number be ' x '

Second natural number be ' x+1

So according to the question,

${x}^{2} + {(x + 1)}^{2} = 61$

let us simplify the expression,

${x}^{2} + {x}^{2} + 1 + 2x - 61 = 0$

$2 {x}^{2} + 2x - 61 = 0$

Divide by 2, we get x

${x}^{2} + x - 30 = 0$

Let us factorize

${x}^{2} + 6x - 5x - 30 = 0$

x(x+6)−5(x+6)=0

(x+6)(x−5)=0

So

(x+6)=0 or (x−5)=0

x=−6 or x=5

∴x=5[ since −6 is not a positive number ]

Hence the first natural number =5

Second natural number =5+1=6

Answer 2

Hope it helps you....✌️✌️