TooEasy

No problem, Guess I'll help...

We let one of our integers be x

and the other integer be y

xy + x + y = 39

We want to maximize the distance between the product of the sum and the product...

(Didn't really understand this part, so my answer could be wrong)
 

So basically, we want our sum to be big, and our product to be big, to get the largest product????

Anyway, to get the largest product, there is a trick...

Try this out yourself:

93 * 95

94 * 94

When evaluating both of them, you will notice that the second expression is greater. This is because the closer the numbers are, the greater the product.

So, when using this theory in our problem, we want the sum and the product of the two integers to be big.

In doing so, we have to think, and for this, I'm pretty sure to just use trial and error.

We have 5 and 4, product 20, sum 9, so total product 180. Pretty good... Except, that they don't sum to 39

We have 19 and 1, product 19, sum 20, so total product = 380. Pretty sure this answer is correct, from what I understand, because the numbers:

19 and 20

Are basically the closest you can go...

So, my answer is 380 as the total product.

Yup, the answer is. I'm so disappointed that the guest can't even do THIS PROBLEM themselves. @Guest, I don't want to be rude or anything, but you really are ruining your future by asking this question here (If you can't do this one ofc)... 

Oh, this looks like a pythag problem, to better understand pythag go to:

Solved this one a few days ago:

Here's the link:

I believe I did say: "might have made a mistake somewhere, why don't you look carefully and read it, instead of just going after the answer"

I answered this same question a few days ago, here's the link:

Here's a tip, go to this link:
 

https://web2.0calc.com/questions/algebra_55659

Beautiful answer by @CPhill

I wrote thought but meant though, I'm too lazy to actually edit it though, so I'll just write this...

You should actually try to do these problems, they would help you a lot in the long run, because quadratics help in many math competitions...

But anyways, I'll still help:

We have:

\(2x^2 + 5x + c = x^2 - 4x\)

When subtracting the x^2 from both sides we get:

\(x^2 + 5x + c = - 4x\)

Add 4x to both sides:

\(x^2+9x + c = 0\)

To have at least one real solution, and the largest... (Hmm, I actually don't know how to do this, but I have intuition!)

I just want you to think about this example for a moment (Solving by intuition again!)

93 * 95 = 8835

94 * 94 = 8836

Do you see something intriguing here? The closer the numbers are together, the greater their product is! (Try it!)

So, we would need two numbers that are close together, possibly equal, and then we will get the greatest product...

We find that these two numbers would have to add to 9, and if we want them to be equal, we use 4.5!

So 4.5 * 4.5 = 20.25, so our constant "c" would be 20.25!

We have to write this as a fraction thought, so:

20 1/4 = 81/4

(Hope my explanation was good, and if there was anything wrong please notify me!)

Honestly, I do think this is right, so could you try again a different way, or maybe they are asking for something more specific... 

@ImCool I'm trying to find your mistake gimme a sec... 

YESSS I FOUND IT (I'm so pleased with my self)

Vieta states that with roots r and s in a simple quadratic:

r+s = negative b

I'm pretty sure this is right based on my knowledge, so that's where you went wrong I think...

We have:

x/x-5 = 4/x-4

Multiply by (x-5) and by (x-4) to both sides

x(x-4) = 4(x-5)

x^2 - 4x = 4x - 20

Put the left side on the right...

x^2 - 8x + 20 = 0

We find factors of 20 that multiply to 20, but add to -8 (we realize factors must be negative...)

Instead, we can just put this into our quadratic formula:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

So we have:

8 +/- root(64 - 80) }/2

8 +/- root(-16) }/2

8 +/- 4i

4 +/- 2i

So the answer should be 4 +/- 2i, note that the 4 is positive...

(If this answer is incorrect please notify me as I would be happy to help) (But you did say this is incorrect, so I don't know what's going wrong here...)