Question

Find the two. Consecutive. Positive integers , such of squares is 365

Answer 1

Let the two consecutive Numbers be x and x+1.
Therefore ,
x² + (x+1)² = 365
x² + x² +1² + 2*x*1 = 365 (because (A+B)² = A² + B²+ 2AB)
or 2x² + 1 + 2x = 365
2x² + 2x = 365 - 1
2x² + 2x = 364
2(x² + x) = 364
or x² + x = 364/2
or x² + x = 182
or x² + x - 182 =0
Now Solve the Quadratic Equation ,
x² + 14x - 13x - 182 = 0
Note : - 13 *14 = 182 , this is because I write 14x - 13x instead of x , so as to solve the quadratic equation .
x (x+14) - 13 (x +14 ) = 0
( x- 13 )(x+14)=0
Therefore , Either x - 13 = 0 or x+14 =0
Since the Consecutive Integers are positive ,
therefore , x-13 = 0
⇒ x =13
hence One of the Positive Integers = 13 ,

therefore other positive integer = x+1 = 13+1 = 14

So the two consecutive positive Integers are 13 and 14 .


Alternate solution


Let The Numbers Be x And x+1

Squares x^2 And (x+1)^2

Sum = 365

Hence

x^2+(x+1)^2 = 365

x^2+x^2+1+2x = 365

2x^2+2x+1 = 365

2x^2+2x-364 = 0

x^2+x-182 = 0

Use Quadratic Formula

-b+Square Root (b^2-4ac) / 2a And -b-Square Root (b^2-4ac) / 2a

-1+Square Root (1+728) / 2 And -1-Square Root (1+728) / 2

-1+27/2 And -1-27/2

13 And -14

As Positive Integers So 13

Other Integer = 13+1 = 14