What 2 numbers have a product of -40 and a sum of -18?

Answer 1

-20 and 2 sorry

Answer 2

alright, for this, you have to create some variables. Let X and Y be the 2 numbers.

so you know that X*Y = -40 and X+Y = -18.

rearrange the first equation by dividing both sides by X, so you get Y = -40/X

Put this value of Y into the second equation:
X + (-40/X) = -18,

multiply both sides by X to get rid of the fraction:
X^2 - 40 = -18X..........bring everything to the left
X^2 + 18X - 40 = 0..........solve the quadratic equation

you could either look for 2 numbers that add to 18 and multiply to -40 (but that is what the question asks for, so if you could do that, then you wouldn't need to do this. so use the quadratic equation:

X = [-18 +- sqrt (18^2 - 4(1)(-40))] / 2
X = [ -18 +- sqrt 484] / 2
X = (-18 +- 22) / 2
X = -40/2 and 4/2
X = -20 and 2

once you have this value of X, put it into any of the 2 equations you have above:

X*Y = -40
(-20) * Y = -40
Y = -40/-20 = 2

X*Y = -40
(2) * Y = -40
Y = -40/2 = -20

X + Y = -18
(-20) + Y = -18
Y = 2

2 + Y = -18
Y = -20




So you get the same values no matter which X you use in the equations. So the 2 numbers are -20 and 2.

Hope that helped you

Answer 3

The numbers are -20 & +2.

Answer 4

It would be -20 and 2

Answer 5

-20*2 = 40
-20+2 = -18

Answer 6

-20 and +2

Answer 7

2 and -20

Answer 8

-20 and 2

Answer 9

(x)(y)= -40
x + y = -18
eq. 2 set = to y: -x-18 = y
plug two into one:
(x)(-x-18)= -40
which becomes:
-x^2 -18 +40 = 0
to solve this quadratic:
[-b +/- sqrt(b^2-4ac)]/(2a)]
a= -1
b= -18
c= 40
when solved : [-b + sqrt(b^2-4ac)]/(2a)]= -20
[-b - sqrt(b^2-4ac)]/(2a)]= 2

Answer 10

Next math test= F

You aren't learning anything and you people aren't helping him by giving him answsers!

Ahhhh.