Find the integers

Question

The sum of the squares of two consecutive odd positive integers is 202. Find the integers.

Answer 1

x and x+2 are the numbers

x²+(x+2)²=202

x²+x²+4x+4=202

2x²+4x+4-202=0

 

2x²+4x-198=0

divide both sides by 2

x²+2x-99=0

factor

(x+11)(x-9)=0

x-9=0

x=9

x+2=11

 

Here's a simple method using arithmetic...

202/2=101, √101=10.049..

the two closest consecutive odd integers are 9 and 11

 

 

 

Answer 2

Let x be the first odd integer, then x + 2 is the next consecutive odd integer.

 

x² + (x + 2)² = 202

x² + x² + 4x + 4 = 202

2x² + 4x + 4 = 202

x² + 2x + 2 = 101

x2 + 2x - 99 = 0

(x +11)(x - 9) = 0,  x = 9 and the next consecutive odd integer is x + 2 = 9 + 2 = 11

 

The two consecutive odd positive integers are 9 and 11

 

Check: 9² + 11² =? 202

81 + 121 = 202