Question

find two consecutive positive integers sum of whose squares is 365

Answer 1

heya here's the answer

let the 2 consecutive integers are x &x+1

sum of their squares is 365

so, x² + (x+1)² = 365

=> x² + x² + 2x + 1 = 365

=> 2x² + 2x = 365 - 1 = 364

=> 2x² + 2x - 364 = 0

=> 2( x² + x - 182) = 0

=> x² + x - 182 = 0

=> x² + 14x - 13x - 182 = 0

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

=> x= -14 or x= 13

since the two numbers are X and X + 1

=> x= 13 & x+1= 13+1= 14

hope it helps u

plz mark as brainlest if it helps

Answer 2

Step-by-step explanation:

AnswEr

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

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

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

$\bigstar$ Finding numbers

First number (x) = 13

Second number (13 + 1) = 14

⠀⠀⠀

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

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