The sum of the squares of two consecutive positive integers is 13.?

Question

What are the integers?

Answer 1

2 squared = 4
3 squared = 9
4 + 9 = 13
therefore 2 and 3 are the two consecutive integers

Answer 2

Let x = the first positive integer
x + 1 = the next positive integer

Equation:

x^2 + (x + 1)^2 = 13
x^2 + (x + 1) (x + 1) = 13
x^2 + x^2 + 2x + 1 = 13
2x^2 + 2x + 1 = 13
2x^2 + 2x + 1 - 13 = 13 - 13
2x^2 + 2x - 12 = 0
2( x^2 + x - 6) = 0
2(x + 3) ( x - 2) = 0
To make the equation true,
x + 3 = 0 or x - 2 = 0
If x + 3 = 0
x = -3 => Rejected because we want positive integer
If x - 2 = 0
x = 2 => Accepted ==> first number
x + 1 = 3 ==> the next number

To check:
x ^2 + ( x+ 1)^ 2 = 13
2^2 + 3^ 2 = 13
4 + 9 = 13
13 = 13

Answer 3

The integers are (2 & 3), because:

2^2 + 3^2 = 13
4 + 9 = 13
13 = 13

2 and 3 are consecutive integers and the sum of their squares is 13.

Answer 4

2 and 3

Answer 5

2 and 3

let x : first integer
x + 1 : next integer
then :
x^2 + (x + 1)^2 = 13
x^2 + (x^2 + 2x + 1) = 13
2x^2 + 2x + 1 = 13
2x^2 + 2x - 12 = 0
2 (x^2 + x - 6) = 0
x^2 + x - 6 = 0
(x + 3)(x - 2) = 0
then either x + 3 = 0 or x - 2 =0
if x + 3 = 0
x = -3 this is not the answer
if x - 2 = 0
x = 2
then x + 1 = 2 + 1 = 3
to check:
2^2 + 3^2 = 13
4 + 9 = 13

Answer 6

let the first positive number be 'x'. SO next number is: x+1
x^2 + (x+1)^2 = 13
x^2 + X^2 + 2x + 1 = 13
2x^2 + 2x = 12
2x(x+1) = 12
x(x+1) = 6

So now the question is simpler. The product of 2 consecutive numbers = 6. So its 2 and 3 since 2x3 = 6

Answer 7

the question can be written as => (x)^2 + (X+2)^2 = 290 => X^2 + X^2+4X+4 = 290 => 2X^2 + 4X+4 = 290 devide both sides by 2 => X^2+2x+2 = 145 => X^2 + 2x-143 = 0 => (X-11)(x+13) = 0 => X-11 = 0 => X=11, hence other number = X+2=13 Ans

Answer 8

yes the numbers would be 2 &3
follow this

assume the 2 nos. x and x+1 since the are consecutive

now (x^2)+(x+1)^2=13
2x^2+1+2x=13
2x^2+2x-12=0
2x^2+6x-4x-12=0
2x(x+3)-4(x+3)=0
(2x-4)(x+3)=0

x=2
put the values and get the answer

Answer 9

let x be the first integer
x+2 the second integer

(x)^2 + (x+2)^2 = 13

x^2 + x^2 + 4x + 4 = 13

2x^2 + 4x - 9 = 0

(2x-3)(x+3) = 0

x = 3/2, -3

Answer 10

x^2 + (x + 1)^2 = 13
x^2 + (x^2 + 2x + 1) = 13
2x^2 + 2x - 12 = 0
x^2 + x - 6 = 0
(x + 3)(x - 2) = 0
x = -3 or 2

since you wanted them positive

ANS : 2 and 3