Question

The product of two consecutive positive integers is 342. We need to find the integers.

Answer 1

Refer to the attachment....

Answer 2

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$\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; Solution :-}}$

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• Let the first positive integer be ' x '
• Let the second positive integer be ' x + 1 '

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⇝ As per the given conditions :-

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$\qquad\tt{:}\longrightarrow\large\textsf{x ( x + 1 ) = 342}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{x² + x = 342}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{x² + x - 342 = 0}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{x² + 19x - 18x - 342 = 0}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{x ( x + 19 ) - 18 ( x + 19 ) = 0}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{( x - 18 ) ( x + 19 ) = 0}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{x - 18 = 0 \; \; or \; \; x + 19 = 0}\\\\\\\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{x = 18 \; \; or \; \; x = - 19}}$

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↭ As the positive integer can't be negative :-

∴ The value of ' x ' will be = 18

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$\boxed{\large\textsf\textcolor{orange}{∴ The first positive integer = 18}}$

$\boxed{\large\textsf\textcolor{orange}{∴ The second positive integer = 19}}$

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$\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; Algebraic \; \; Formulas :-}}$

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$\qquad\tt{:}\longrightarrow\large\textsf{( a + b )² = a² + b² + 2ab}$

$\qquad\tt{:}\longrightarrow\large\textsf{( a - b )² = a² + b² - 2ab}$

$\qquad\tt{:}\longrightarrow\large\textsf{a² - b² = ( a + b ) ( a - b )}$

$\qquad\tt{:}\longrightarrow\large\textsf{a² + b² = ( a + b )² - 2ab}$

$\qquad\tt{:}\longrightarrow\large\textsf{a³ + b³ = ( a + b ) ( a² - 2ab + b² )}$

$\qquad\tt{:}\longrightarrow\large\textsf{a³ - b³ = ( a - b ) ( a² + ab + b² )}$

$\qquad\tt{:}\longrightarrow\large\textsf{( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ac }$

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