Math, asked by prathammittal357, 6 months ago

The product of two positive integer is 12. If their sum added to the sum of their square is 32. find the numbers

Answers

Answered by AlluringNightingale
1

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 .

Thus ,

Required integers are 3 and 4 .

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