Verified answer

Answer:

117

Step-by-step explanation:

Given :

• The difference of two positive whole numbers is 3.

. The sum of their squares is 117.

To find :

• The two positive numbers.

Solution :

Let the required two positive numbers be X and Y

Let X > Y

Given that

The difference of two positive whole

numbers

= 3

=> X - Y = 3

-(1)

and

The sum of their squares = 117

=> x² + y² = 117-------- -(2)

We know that

(a+b)² = a²+2ab+b²

=> (X-Y)²=X²-2XY+Y²

=> 3² = 117-2XY

=> 9-117-2XY

=> -108 = -2XY

=> XY = -108/-2

=> XY = 54

-(3)

--(4)

We know that

(a+b)² = (a-b)² +4ab

=> (X+Y)² = 3²+4(54)

=> (X+Y)² = 9+216

=> (X+Y)² = 225

=> X+Y = √225

=> X+Y = +15

Therefore, X+Y = 15

Since, X and Y can not be negative.

On adding (1)&(4)

X-Y = 3

X+Y = 15

(+)

=> X = 18/2

=> X = 9

On substituting the value of X in (4) then

9+Y = 15

=> Y = 15-9

=> Y = 6

Therefore, X = 9 and Y=6

Answer :

◆ The required two positive whole numbers are 9 and 6.

Check :

The two numbers = 9 and 6

Their difference = 9-6 = 3

The sum of their squares

= 9²+6²

= 81+36

= 117

Verified the given relations in the given problem.

Used formulae:

◆ (a+b)² = a²+2ab+b²

◆ (a+b)² = (a-b)² + 4ab