Question

Find the two consecutive positive integer, sum of whose squares is365

Answer 1

Answer:

182, 183

Step-by-step explanation:

let one no. be x

let other no. be x+1

A/Q  x+(x+1)=365

        2x+1=365

        2x=364

         x=182

then further put the value 182 on the place of x and u can even check your answer.

hope, you like it

Answer 2

$\sf\large\underline{Let:}$

$\rm{\implies The\:2\: consecutive\: integer\:be\:x\:and\:x+1}$

$\sf\large\underline{To\: Find:}$

$\rm{\implies The\: consecutive\: integer=?}$

$\sf\large\underline{Solution:}$

$\rm{\implies (x)^2+(x+1)^2=365}$

$\rm{\implies x^2+x^2+2x+1=265}$

$\rm{\implies 2x^2+2x+1=365}$

$\rm{\implies 2x^2+2x=365-1}$

$\rm{\implies 2x^2+2x=364}$

$\rm{\implies 2x^2+2x-364=0}$

• Dividing by 2 on both sides here]

$\rm{\implies x^2+x-182=0}$

• Here splitting the middle term here]

$\rm{\implies x^2+14x-13x-182=0}$

$\rm{\implies x(x+14)-13(x+14)=0}$

$\rm{\implies (x-13)(x+14)=0}$

$\rm{\implies \therefore\:x=13\:and\:-14}$

• Here negative value can't be the number of integer so we take x=13]

$\sf\large{Hence,}$

$\bf{\implies First\:_{(integer)}=x=13}$

$\bf{\implies Second\:_{(integer)}=x+1=13+1=14}$