Question

find the two natural numbers which differ by 5 and the sum of whose squares is 97

Answer 1

Let x and x+5 be the numbers. 

x² + (x+5)² = 97 

Solve for x 

x²+ x² +10x + 25 = 97 

2x² + 10x - 72 = 0 

Simplify by 2 

x² + 5x - 36 = 0 

(x-4)(x+9)=0 

Since x is positive, only x-4 can be 0. 

x = 4 and x+5 = 9 



Check: 16 + 81 = 97

Answer 2

Answer:is 9 and4

Step-by-step explanation:4 and 9 are the required numbers

Step-by-step explanation:

Let x and x + 5 be the numbers.

x² + ( x+5 )² = 97

x² + x² + 10x + 25 = 97

2x² + 10x - 72 = 0

Dividing all through by 2, we get

x² + 5x - 36 = 0

(x-4) (x+9) = 0

Since x is positive, only (x-4) can be the zero.

So, x = 4 and x = 5 + 4 = 9

Therefore, the two natural numbers are 4 and 9