The product of two positive integer is 12. If their sum added to the sum of their square is 32. find the numbers
Answers
Answer :
3 , 4
Solution :
Let the two required positive integers be a and b .
According to the question , the product of the two integers is 12 .
Thus ,
ab = 12 -------(1)
Also ,
It is given that , their sum added to the sum of their squares is 32 .
Thus ,
(a² + b²) + (a + b) = 32 -----(2)
Now ,
Adding 2ab on both the sides of eq-(2) , we get ;
=> (a² + b² + 2ab) + (a + b) = 32 + 2ab
=> (a + b)² + (a + b) = 32 + 2•12
=> (a + b)² + (a + b) = 32 + 24
=> (a + b)² + (a + b) = 56 -----(3)
Now ,
Putting a + b = y , eq-(3) will reduce to
y² + y = 56
=> y² + y - 56 = 0
=> y² + 8y - 7y - 56 = 0
=> y(y + 8) - 7(y + 8) = 0
=> (y + 8)(y - 7) = 0
=> y = -8 , 7
=> y = 7 (appropriate value)
[ Note : y = -8 is rejected value because a and b are positive integers , thus their sum y = a + b must be a positive integer but y = a + b = -8 is a negative integer hence it is rejected . ]
Hence ,
y = 7
ie . a + b = 7 --------(7)
=> b = 7 - a --------(8)
Now ,
Putting b = 7 - a in eq-(1) , we get ;
=> ab = 12
=> a(7 - a) = 12
=> 7a - a² = 12
=> a² - 7a + 12 = 0
=> a² - 4a - 3a + 12 = 0
=> a(a - 4) - 3(a - 4) = 0
=> (a - 4)(a - 3) = 0
=> a = 4 , 3
• Case 1 :
If a = 4 , then
=> b = 12/a [ Using eq-(1) ]
=> b = 12/4
=> b = 3
• Case 2 :
If a = 3 , then
=> b = 12/a [ Using eq-(1) ]
=> b = 12/3
=> b = 4
Clearly ,
In both the cases , we got the same pair of integers .