Question

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Answer 1

Let's consider that two consecutive positive integers are x & (x + 1).

$\begin{gathered}{\underline{\sf{\bigstar\: According \ to \ the \ given \ Question :}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf x^2 + \Big(x + 1 \Big)^2 = 365 \\\\\\:\implies\sf x^2 + x^2 + 1 + 2x = 365 \\\\\\:\implies\sf 2x^2 + 2x^2 = 365 - 1 \\\\\\:\implies\sf 2x^2 + 2x- 364 = 0 \qquad \bigg\lgroup\sf Taking \ 2 \ common \bigg\rgroup\\\\\\:\implies\sf x^2 + x - 182 = 0 \\\\\\\qquad\qquad\underline{\sf{\purple{\: Using \ splitting \ the \ Middle \ term \ method \ :}}}\\\\\\:\implies\sf x^2 + 14 x - 13x - 182 = 0\\\\\\:\implies\sf x( x + 14) - 13(x + 14) = 0\\\\\\:\implies\sf\pink{(x - 13) (x + 14) = 0}\\\\\\:\implies\sf x - 13 = 0 \\\\\\:\implies\boxed{\rm{\blue{\: x = 13}}} \\\\\\:\implies\sf x + 14 = 0 \\\\\\:\implies\boxed{\rm{\blue{\: x = -14}}}\end{gathered}$

$\bigstar★ Finding numbers$

First number (x) = 13

Second number (13 + 1) = 14[/tex]

⠀⠀⠀

$\begin{gathered}\therefore\underline{\textsf{Two positive consecutive numbers are \textbf{13 \& 14}}}. \\ \end{gathered}$

Answer 2

thanks for telling this bro