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## Answers

*let the no. be x and x+1*

*then we have x²+(x+1)²=365*

*2x²+2x-364=0*

*x²+x-182=0*

*x²+14x-13x-182=0*

*(x+14)(x-13)=0*

*That gives x=13 Avoiding negative value*

*one number is 13 and other one 14.*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 .

Hope this helps You !!

Thanks Cheers !!

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 .

Hope this helps You !!

Thanks Cheers !!