Question

Write a pair of integers whose product is -36 and whose difference is 15

Answer 1

Answer:

1st pair : 12 and -3

2nd pair : 3 and -12

Solution:

Let the two integers be x and y such that x > y .

It is given that ,

The product of the integers is -36 and their difference is 15 .

Thus,

x•y = -36 -------(1)

x - y = 15 ------(2)

Now,

=> x•y = -36

=> (15+y)•y = -36. { using eq-(2) }

=> 15y + y² = 36

=> y² + 15y + 36 = 0

=> y² + 12y + 3y + 36 = 0

=> y(y + 12) + 3(y + 12) = 0

=> (y + 12)(y + 3) = 0

=> y = - 3 , -12

Now,

Using eq-(1)

If y = -3

then x = -36/y = -36/-3 = 12

If y = -12

then x = -36/y = -36/-12 = 3

Hence,

Two such pairs of integers are possible :

1st pair : 12 and -3

2nd pair : 3 and -12

Verification:

Case1 : x = 12 and y = -3

Product = x•y = 12•(-3) = -36

Difference = x - y = 12 - (-3) = 12 + 3 = 15

Case2 : x = 3 and y = -12

Product = x•y = 3•(-12) = -26

Difference = x - y = 3 - (-12) = 3 + 12 = 15

Clearly,

In both the cases , the product of the integers is -36 and their difference is 15 .

Answer 2

Answer: 12 and -3

 Explanation: Let the two integers be x and y such that x > y.

 Then, the product of the integers is -36 and their difference is 15.

Therefore,

x•y = -36 -------(1)

x - y = 15 ------(2)

Now,

x•y = -36

(15+y)•y = -36. { using eq-(2) }

15y + y² = 36

y² + 15y + 36 = 0

y² + 12y + 3y + 36 = 0

y(y + 12) + 3(y + 12) = 0

(y + 12)(y + 3) = 0

 y = - 3 , -12