CPhill

Suppose we have the quadratic....... ax^2 + bx + c

This technique will work as long as we can find another factorization of ac where the factors sum to "b"

To see why, let's suppose ac = n

Now, let us suppose that there are two other factors of n, say q and r, such that q + r = b....then...

 (a must equal q*m)  and (c must equal r/m)  because (qm) * (r/m) = q*r = n

So, we have..........

ax^2 + bx + c =

(q*m)x^2 + (q + r) x  + (r/m) =

(q*m)x^2 + qx  + rx + (r/m) =

qmx(x + 1/m) + r (x + 1/m) =

(qmx + r) ( x + 1/m)

Which shows that this type of polynomial can be factored as long as ac - i.e., n - has factors q,r such that q + r = b

Ah....I see what you mean, reinout!!!  Sorry, I misread it !!!

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Let's try this....maybe the bottle of pills costs $80........???

So, 80/30 = 8/3  = 2 + 2/3    =  about  $2.66 per pill

To borrow a term from ND ....that is "spendy!!!"  (my new Word of the Day!!!)

BTW....I agree with rosala.....definitely a "museum" piece  !!!

We would have to know the area and the size of a tile !!!

I actually don't know how to get the inverse, mathematically......maybe some other people on the site know a trick or two and could show you....!!!

Strictly speaking, we would have to have a restricted domain on the original function since it isn't one-to-one.....

However, here's what WolframAlpha has .......(the red graph is the original and the blue one is the inverse)

We would want to write it as:

f^-1(x) = (x+2)/2 ±(1/2)√(x)√(x+4)

Mmmmm....that's a variation on a theme of one of the great "open" questions in mathematics...!!!

It's known as The Goldbach Conjecture and it states that every positive even integer can be expressed as the sum of two primes (maybe in more than one way, too. For instance, 24 = 19+5 = 13+11). As yet, it's never been proved.........

As to your specific question, I'm not sure that's been proved, either......but maybe someone else on this site would have more information!!!!

A mathematician by the name of Lagrange did prove that every "natural" number could be written as a sum of four squares.........

Here's a link to Goldbach' Conjecture which you might find of interest :

ND......I don't find the on-site grapher very intuitive.....  I've never been able to figure it out.....Desmos is DEFINITELY the way to go!!  It produces a way "cleaner" graph, too.

Another site.....WolframAlpha can produce some nice graphs, too......I've copied some of these into the posts.....sometimes they stay there.....sometimes they don't......I don't remember if I've tried the Tiny Pic thing with it, or not.....

Anyway, that Wolfram site has a lot of cool math capabilities....you should check it out!! Just type in what you want to do and it will take you to the correct page....for example....typing in "algebra" brings up a whole array of stuff you can do.....I've bookmarked some of the pages I use most often......

Here's the link :

It's called "Desmos Graphing Calculator," ND......easy to use.......here's the link :

We have the equation of a line, here.  Let's do a bit of rearranging.......we have

17x+21y+11=0     subtract 17x and 11 from both sides...this gives

21y = -17x -11     divide by 21  on both sides

y = (-17/21) - 11/21

This is a line with a slope of (-17/21)  and a "y intercept" of -11/21

Here's the graph: