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

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Let the numbers be x and y . As their sum is 4 and their product is -96.

=> x + y = 4 and xy = -96

x + y = 4

=> x = 4 - y

substitute this in xy = -96

=> (4 - y)y = -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 or -12

x = -12 or 8

**Therefore the numbers are 8 and -12.**

Let a and b be the number such that their sum is 4 and their product is -96.

Then we will write:

a + b = 4 ...............(1)

a*b = -96 .............(2)

Now we have a system of two equations and two variables.

We will use the substitution method to solve.

We will wrtie (1).

a + b = 4

==> a = 4 - b

Now we will substitute in (2).

a*b = -96

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

==> 4b - b^2 = -96

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

Now we will factor.

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

==> b1 = 12 ==> a1 = 4-12 = -8

==> b2= -8 ==> a2= 4-(-8) == 12

**Then, the numbers are 12 and -8.**

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**

Since the sum of 2 numbers is 4, we assume x and 4-x .

Since the product of these mumbers is -96, x(4-x) = -96.

Or x(x-4) = 96

Therefore x^2-4x-96 = 0.

x^2-12x+8x-96 = 0.

x(x-12) +8(x-12) = 0.

(x-12)(x+8) = 0.

x-12 = 0 or x+8= 0.

x =12 , or x=-8.

Therefore the solutions of the give equation are x = 12 and x = -8.