Question

Find two consecutive positive integers, sum of whose squares is 613.

Answer 1

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Here is your answer!!!

let the numbers be x and x+1

A.T.Q,
Let the integers be x and x+1
Given that sum of consecutive positive integers is 613.
Now,Acc. to statement:

=>x^2+(x+1)^2=613

=> x^2+x2+1+2x=613

=> 2x^2+2x=612

=> x^2 + (x + 1)^2 = 613

=> x^2 + x^2 + 1 + 2x = 612

=> 2x^2 + 2x = 612

=> 2x^2 + 2x - 612 = 0

=> 2(x^2 + x - 306) = 0

=> x^2 + 18x - 17x - 306 = 0

=> x(x + 18) - 17(x + 18) = 0

= > (x + 18)(x - 17) = 0

=>x=17and−18​

Since, X is a positive integer so we reject x = -18.
Hence,x=17 and x+1 =18.

So , The two consecutive positive integers are 17 & 18.

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Answer 2

HLO MATE !!!!
Simran Here !!!

HERE IS UR ANSWER _____________
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$Let \: the \: integers \: be \: x \: and \: x + 1 \\ \\ Given \: that \: sum \: of \: consecutive \: positive \: integers \: is \: 613 \\ \\ Now, \: Acc \: to \: statement : \\ \\ = > x {}^{2} + (x + 1) {}^{2} = 613 \\ \\ = > x {}^{2} + x {}^{2} + 1 + 2x = 613 \\ \\ = > 2x {}^{2} + 2x = 612 \\ \\ = > 2x {}^{2} + 2x - 612 = 0 \\ \\ = > 2(x ^{2} + x - 306) = 0 \\ \\ = > x {}^{2} + 18x - 17x - 306 = 0 \\ \\ = > x(x + 18) - 17(x + 18) = 0 \\ \\ = > (x + 18)(x - 17) = 0 \\ \\ = > x = 17 \: and \: - 18$

So , The two consecutive positive integers are 17 & 18.....
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