# Numbers .What are numbers whose sum is 4 and the product is -96 ?

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### 3 Answers

Let the numbers be A and B

The sum of the numbers is 4

=> A + B = 4

=> A = 4 - B

The product is -96

=> A*B = -96

=> (4 - B)*B = -96

=> 4B - B^2 = -96

=> B^2 - 4B - 96 = 0

=> B^2 - 12B + 8B - 96 = 0

=> B(B - 12) + 8(B - 12) = 0

=> (B + 8)(B - 12) = 0

=> B = -8 and B = 12

A = 12 and -8

**The two numbers are 12 and -8**

The sum of two numbers is 4 and their product is -96. If one of the numbers is taken to be x and the other is y, x + y = 4 and x*y = 96

x + y = 4 gives x = 4 - y

Substitute this in x*y = -96

(4 - y)*y = -96

4y - y^2 = -96

y^2 - 4y - 96 = 0

y^2 - 12y + 8y - 96 = 0

y(y - 12) + 8(y - 12) = 0

(y + 8)(y - 12) = 0

y = -8 and y = 12

The two numbers are -8 and 12.

We'll use the sum and the product to form the quadratic equation, whose root are representing the requested numbers.

S = 4 and P = -96

W'll form the quadratic:

x^2 - 4x - 96 = 0

We'll apply the quadratic formula:

x1 = [4 + sqrt(16 + 384)]/2

x1 = (4+20)/2

x1 = 12

x2 = (4-20)/2

x2 = -8

**The numbers whose sum is 4 and product is -96 are: x1 = 12 and x2 = -8**