Question

Find two natural numbers which differ by 3 and whose squares have the sum 117.

Answer 1

SOLUTION :

Given : The two natural numbers differ by 3  

Let the one natural number be x and other natural number be (x -3)

A.T.Q

(x)² + (x - 3)² = 117

x² + x² + 9 - 6x -117 = 0

[(a - b)² = a² + b² - 2ab ]

2x² -6x -108 = 0

2(x² - 3x - 54) = 0

x² - 3x - 54 = 0

x² - 9x + 6x - 54 = 0

[By middle term splitting ]

x(x - 9) + 6(x - 9) = 0

(x - 9)(x + 6) = 0

(x - 9) = 0  or (x + 6) = 0

x = 9  or x = - 6  

Since, x is a natural number, so x ≠ - 6

Therefore, x = 9  

First natural number = x = 9

Second natural number = (x -3) = 9 -3 = 6  

Hence, the two natural  numbers are 6 and 9 .

HOPE THIS  ANSWER WILL HELP YOU…..

Answer 2

Step-by-step explanation:

Ans : 6, 9

Hint :

Let the numbers be x and x - 3. Then, x

2

+(x−3)

2

=117.