The difference of two positive whole numbers is 3 and the sum of their square is 117;by calculating,let us write the two numbers.
Answers
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
=> 9-117 = -2XY
=> -108 = -2XY
=> XY = -108/-2
=> XY = 54 -------------(3)
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 --------(4)
Since, X and Y can not be negative.
On adding (1)&(4)
X-Y = 3
X+Y = 15
(+)
________
2X+0 = 18
________
=> 2X = 18
=> 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
Answer:
9 - 6 = 3 and 9 square is 81, 6 square is 36 so by adding them both we get 81+36=117
Step-by-step explanation:
81+36=117