Question

How do I find two integers whose sum is -8 and product is -48?

Answer 1

4 and -12

Explanation:

You've got those two equations:
a*b=-48
a+b=-8
From the second equation we get a=-8-b, and then we put -8-b instead of 'a' in the first equation:
`(-8-b)*b=-48`
`-8b-b^2=-48`
`b^2+8b-48=0`
`D=64+4*48=256=16^2>0`
`b=(-8+16)/(2)=4` or `b=(-8-16)/(2)=-12`
From here as a=-8-b, then if `b=4`, then `a=-8-4=-12` and if `b=-12`, then `a=-8-(-12)=4` So the numbers are 4 and -12.

Answer 2

`-12" and "4`

Explanation:

`"since the product is negative then one number must be"`
`"positive and the other negative"`

`"let the numbers be "x" and "y" then"`

`x+y=-8to(1)`

`xy=-48to(2)`

`"from equation "(1)color(white)(x)y=-8-xto(3)`

`"substitute "y=-8-x" in equation "(2)`

`x(-8-x)=-48`

`rArr-8x-x^2=-48`

`"this is a quadratic equation so express in standard form"`

`x^2+8x-48=0larrcolor(blue)"in standard form"`

`"the factors of - 48 which sum to + 8 are +12 and - 4"`

`rArr(x+12)(x-4)=0`

`"equate each factor to zero and solve for x"`

`x+12=0rArrx=-12`

`x-4=0rArrx=4`

`"substitute these values into equation "(3)`

`x=-12rArry=-8+12=4`

`x=4rArry=-8-4=-12`

`rArr"the two integers are "-12" and "+4`