Question

find two consecutive positive integers, sum of whose square is 365​

Answer 1

Answer:

there is difference of 1 in consecutive positive integers. Let's first integer = x. so, second integer = x+1. also given that, sum of squares=365(first number)square +(second number)square =365

Answer 2

⤤ Given :-

• sum of two consecutive positive integers of square is 365

⤤ To find :-

• Two consecutive positive integers.

⤤ Solution :-

Let the first consecutive positive integer be x

and second consecutive positive integer be x + 1

⸕ According to the question,

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

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

$\rm\longrightarrow\:2{x}^{2}+2x+1-365=0$

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

$\rm\longrightarrow\:2({x}^{2}+x-182)=0$

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

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

$\rm\longrightarrow\:x(x+14)-13(x+14)=0$

$\rm\longrightarrow\:(x+14)(x-13)=0$

Now,

$\rm\longrightarrow\:x+14=0$

$\rm\longrightarrow\:x=0-14$

$\rm\longrightarrow\:x=-14$

Then,

$\rm\longrightarrow\:x-13=0$

$\rm\longrightarrow\:x=0+13$

$\rm\longrightarrow\:x=13$

Since,the integers are positive so, x will be 13

and

➝ x + 1

➝ 13 + 1

➝ 14

Hence,the two consecutive positive integers will be 13 and 14.