Question

The square of two consecutive integers differ by 13 ,then the largest integer is

Answer 1

let the first no.=x
second no.=x+1

x^2-(x+1)^2=13
x^2-(x^2+2x+1^2)=13
x^2-x^2+2x+1=13
2x+1=13
2x=13-1
2x=12
×=12/2
x=6

largest no.= x+1
=6+1=7

Answer 2

Concept:-

It might resemble a word or a number representation of the quantity's arithmetic value. It could resemble a word or a number that represents the numerical value of the quantity. It could have the appearance of a word or a number that denotes the quantity's numerical value.

Given:-

We have been given that the square of two consecutive integers differs by $13$.

Find:-

We need to find the largest integer of the square of two consecutive integers differs by $13$.

Solution:-

Let the integers be $x$ and $x + 1$.

Given $(x+1)^2-x^2=13$. According to the given condition,

$x^2 +2x+1-x^2=13\\2x+1=13\\2x=13-1\\2x=12\\x=\frac{12}{2}\\=6$

The largest integer is $x+1=6+1=7$

Hence the largest integer is $7$.

#SPJ2