the difference of two positive numbers is six. Their product is 223 less than the sum of their squares. what are the two numbers
Answers
Step-by-step explanation:
Given :-
The difference of two positive numbers is six. Their product is 223 less than the sum of their squares.
To find :-
What are the two numbers ?
Solution :-
Let the two numbers be X and Y (X>Y)
Given that
The difference of two positive numbers = 6
=> X-Y = 6
=> X = Y+6 -----------(1)
Their Product = X×Y = XY
Square of X = X²
Square of Y = Y²
Given that
Their product is 223 less than the sum of their squares.
=> XY = (X²+Y²)-223
=> (Y+6)(Y) = (Y+6)²+Y²-223
=> Y²+6Y = Y²+2(Y)(6)+6²+Y²-223
Since (a+b)² = a²+2ab+b²
=> Y²+6Y = Y²+12Y+36+Y²-223
=> Y²+6Y = 2Y²+12Y- 187
=> 2Y²+12Y-187-Y²-6Y = 0
=> Y²+6Y-187 = 0
=> Y² + 17Y-11Y-187 = 0
=> Y(Y+17)-11(Y+17) = 0
=> (Y+17)(Y-11) = 0
=> Y+17 = 0 or Y-11 = 0
=> Y = -17 or Y = 11
Given numbers are positive then
Y = 11
On Substituting the value of Y in (1)
=> X = 11+6
=> X = 17
Therefore, X = 17 and Y = 11
The numbers are 17 and 11
Answer:-
The required numbers for the given problem are 17 and 11
Check :-
The two numbers = 17 and 11
Their difference = 17-11 = 6
their product = 17×11 = 187
The square of 17 = 17² = 17×17 = 289
The square of 11 = 11² = 11×11 = 121
Sum of their squares = 289+ 121 = 410
=> 410 -223
=> 187
Their product is 223 less than the sum of their squares.
Verified the given relations in the given problem.
Used formulae:-
- (a+b)² = a²+2ab+b²