Question

Find the two consecutive integers sum of whose square is 365

Answer 1

HeYa❤️...

Answer:

Let the first consecutive number be (x).

And second number be (x+1).

A.T.Q,

$(x) {}^{2} + (x + 1) {}^{2} = 365 \\ \\ x {}^{2} + x {}^{2} + 1 + 2x = 365 \\ \\ 2x {}^{2} + 2x + 1 - 365 = 0 \\ \\ 2x {}^{2} + 2x - 364 = 0$

By factorisation method,

$x = \frac{ - b + \sqrt{b {}^{2} - 4ac } }{2a} \\ \\ x = \frac{ - 2 + \sqrt{} {b}^{2} - 4ac }{2 \times 2} \\ \\ x = \frac{ - 2 + \sqrt{} 4 - 2912}{4} \\ \\ x = \frac{ - 2 - 53.92}{4} \\ \\ x = \frac{ - 55.92}{4} \\ \\ x = 54.42 \: or \: x = - 54.42$

Hence, the two consecutive numbers are 54.42 and 54.42+1 = 55.42

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