Question

$\huge \fbox \red{❥ Question}$

Find two consecutive positive integers, sum of whose squares is 365.​

Answer 1

${\huge{\rm{\underline{\underline{Question:-}}}}}$

→ Find two consecutive positive integers, sum of whose squares is 365.​

${\huge{\rm{\underline{\underline{Solution:-}}}}}$

→ Let the first consecutive integer be ( x )

  second integer will be ( x + 1 )

→ According to the question !!

$\sf \longrightarrow x^{2} + (x+1)^{2} = 365\\\\\longrightarrow x^{2} + x^{2} + 2x + 1 = 365\\\\\longrightarrow x^{2} + 2x - 364 = 0 \\\\\longrightarrow x^{2} + x - 182 = 0 \\\\Let\;us\;split\;the\;middle\;term\\\\\longrightarrow x^{2} + 14x - 13x -182 = 0 \\\\\longrightarrow x(x+14) - 13(x+14) = 0 \\\\\longrightarrow (x+14)(x-13) = 0 \\\\x = -14 \;, \; x = 13 \\\\( negative\;root\;being\;rejected)$

${\huge{\rm{\underline{\underline{Hence,\;required\;numbers\;are\;13\;and\;14}}}}}$

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All Done !! :D