Question

Two numbers have a difference of 16 . Find the numbers if the result of adding their sum and their product is a minimum. Please show all your work and explain the steps. Thank you!

Answer 1

Call the numbers $x$, $y$ and suppose $x>y$
Their difference is $16$ :$x-y=16\implies\,y=x-16$

Their sum is $x+y$ and product is $xy$
Adding sum and product gives $x+y+xy$

Substitute the $y$ value and get a quadratic in $x$$x+y+xy=x+x-16+x(x-16)\\=x^2-14x-16$
Minimum value occurs at vertex and the $x$ coordinate of vertex is $x=-\frac{b}{2a}=-\frac{-14}{2}=7$

Plugging $x$ in earlier equation gives $y=7-16=-9$
Therefore the required numbers are $7$ and $-9$

Answer 2

Answer:

The numbers are 7 and -9.

Step-by-step explanation:

let x represent the first number.

let y represent the second number. it is not x + something because the question does not say the second number is consecutive or anything like that.

x-y = 16 (since the difference of x and y is 16)

isolate for y (but you can isolate for x if you want. i like doing it for y better)

x = y +16

x - 16 = y

the equation for the question should be:

x + y + xy = 0

the "x + y" comes from the sum, and "xy" comes from making the product. the question asks to add them together. we want to factor in the end to find x, and the only way to do that is if we have zero on the other side of the equation.

substitute the new y value into the equation

x + (x - 16) + x(x - 16) = 0

collect like terms and use distributive property

x + x - 16 + x^2 -16x = 0

2x - 16 + x^2 -16x = 0

x^2 -14x -16 = 0

you now have a quadratic. to find x, convert the equation into vertex form (aka completing the square). we cannot factor in any other way (technically we can, it just takes longer).

(x^2-14x) - 16 = 0

(x^2 - 14x + 7^2 - 7^2) - 16 = 0

(x^2 - 14x + 49) - 49 - 16 = 0

(x - 7)^2 - 65 = 0

find x algebraically

x - 7 = 0

x = 7

now that we found x, we can sub it into the y equation to find y

y = x - 16

y = 7 - 16

y = -9

therefore, the numbers are 7 and -9.

hope this helps :)