What is the minimum possible product of two numbers that differ by $8$ $8$ ?

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What is the minimum possible product of two numbers that differ by $8$ $8$ ?

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Answer 1 · 2016-11-06T08:15:39

The minimum product is $- 16$ $-16$ and the two numbers are $- 4$ $-4$ and $4$ $4$

Explanation:

Let the two numbers be $x$ $x$ and $x + 8$ $x+8$ .

Then their product is:

$f (x) = x (x + 8)$ $f(x) = x(x+8)$

$f (x) = x^{2} + 8 x$ $color(white)(f(x)) = x^2+8x$

$f (x) = x^{2} + 8 x + 16 - 16$ $color(white)(f(x)) = x^2+8x+16-16$

$f (x) = {(x + 4)}^{2} - 16$ $color(white)(f(x)) = (x+4)^2-16$

For any Real value of $x$ $x$ we will have ${(x + 4)}^{2} \geq 0$ $(x+4)^2 >= 0$

Hence $f (x)$ $f(x)$ attains its minimum value $- 16$ $-16$ when ${(x + 4)}^{2} = 0$ $(x+4)^2 = 0$

That is when $x = - 4$ $x = -4$

So the minimum product is $- 16$ $-16$ and the two numbers are $- 4$ $-4$ and $4$ $4$

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What is the minimum possible product of two numbers that differ by $8$ $8$ ?

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