Find two numbers that have a sum of -2 and a product of -48.

Answer 1

Here's a good start (I hope)

I'm going to assume it's not enough to run a few numbers through your head (the factors of 48, for example) and trial and error your way to a solution.

You (your teacher) want to solve it algebraically . . . So here goes . . .

Two numbers add to -2 means x + y = -2

And multiply to -48 means xy (x times y) = -48

Choose either equation (x+y=-2 or xy=-48) and solve for one of the variables

I prefer keeping x+y=-2 the way it is

And solving the other equation for x

You get x=-48/y (-48 divided by y)

Now a little substitution . . . if x=-48/y, you can substitute that into the first equation for the x term

So, x+y=-2 becomes -48/y + y=-2

How do you get rid of the "y" in the denominator?

Multiply "everything" by y (the -48/y, the y, and the -2 on the other side of the equation)

You get -48 + y^2 = -2y (-48 + y-squared = -2y)

"Bring over" the -2y (by adding 2y to both sides)

-48 +y^2 + 2y = 0

Rearrange the terms

y^2 +2y -48 = 0

Now you can solve this obvious quadratic equation in the form ax^2 + bx + c (or ay^2 + by + c = 0) using the Quadratic Formula . . . but first . . . check to see if you can factor (Hint: Don't use the quadratic formula)

Here's where I'll leave it to you

Answer 2

We have 2 unknowns, so we need 2 equations to solve this.

Let x = first number

y = second number

Eq. 1: x + y = -2---->x = -2 - y

Eq. 2: xy = -48

Putting the value of "x" from Eq. 1 into Eq. 2 we can solve for "y":

xy = -48 Eq. 1

(-2 - y)y = -48 substitute -2 - y for x

-2y - y² = -48 simplify

y² + 2y - 48 = 0 add 2y and y² to both sides

(y-6)(y+8) = 0 factor

y = 6 y = -8 solve y so that the equation would equal zero

Using x = -2 -y, solve for x for each y value:

For y = 6: x = -2 -6---->x = -8

For y = -8: x = -2 - (-8)---->x = -2 + 8---->x = 6

Therefore, the 2 numbers would be 6 and -8.

Hopr this helps!

Answer 3

The answer would be -8 and positive 6

-8 times 6=-48

and

-8+6=-2

I hope that helps!