Question

find 2 consecutive positive integers sum of whose square is 365​

Answer 1

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$\large\pink{\underline{\underline{\maltese{Given:-}\maltese}}}$

• 2 consecutive positive integers
• sum of their square is 365.

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• the two numbers.

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• Mid term factorisation

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Let the two numbers be x and (x+1) .

Sum of their square is 365.

x² + (x+1)² = 365

=> x² + x² + 2x + 1 = 365

=> 2x² + 2x = 365-1

=> 2x² + 2x = 364

=> 2x² + 2x - 364 = 0

=> 2 ( x² + x - 182 ) = 0

=> x² + x - 182 = 0

=> x² + 14x - 13x - 182 = 0

=> x ( x + 14 ) - 13 ( x - 14 ) = 0

=> ( x + 14 ) ( x - 13 ) = 0

=> x+14= 0 or x-13= 0

=> x = -14 or x = 13

Since both are positive integers ,

x => 13

x+1 = 13+1 => 14

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Answer 2

Sum of their square is 365.

x² + (x+1)² = 365

=> x² + x² + 2x + 1 = 365

=> 2x² + 2x = 365-1

=> 2x² + 2x = 364

=> 2x² + 2x - 364 = 0

=> 2 ( x² + x - 182 ) = 0

=> x² + x - 182 = 0

=> x² + 14x - 13x - 182 = 0

=> x ( x + 14 ) - 13 ( x - 14 ) = 0

=> ( x + 14 ) ( x - 13 ) = 0

=> x+14= 0 or x-13= 0

=> x = -14 or x = 13

Since both are positive integers ,

x => 13

x+1 = 13+1 => 14