Sorry for this not being entirely relevant to the thread but it involves the quadratic formula and stuff so I'll post it here, instead of bumping that other maths thread from last year. Also sorry for probably being awful at explaining things well over text. This was originally going to be a question asking for help but I worked it out in the middle of writing the post, heh. (but then more questions came up but then I worked them out but then...)

Someone from the year below me asked for some help with his maths homework. His question was:

The sum of two of a rectangle's sides is S, the area is A. Prove that S² ≥ 4A. (Suggestion: The rectangle's sides are x and S-x)

Since the homework itself involved the quadratic formula, I assume there's some way to prove it using that. However the only way I could get an answer at the time was by doing this

x(S-x) = A

-x² + Sx = A

x² - Sx = -A

(x - S/2)² - S²/4 = -A

4(x - S/2)² - S² = -4A

His class haven't done complex numbers so (x - S/2)² ≥ 0, and so S² which is being taken away from this has to be greater than or equal to 4A for it to make -4A.

I'm sure there's some other way to work it out though, since I remember doing the same homework last year. I then realised the rest of the questions on that homework sheet were proving that a quadratic equation did/n't have real roots by working out b² - 4ac

This (much simpler!) way obviously works:

x² - Sx + A = 0

b² - 4ac ≥ 0

S² - 4A ≥ 0 (I realise that it's (-S)² but I usually just skip that step if it's negative)

S² ≥ 4A

This was the reason I was going to make this post in the first place since I didn't see why it needed real roots as I had no idea what the roots were in the context of the rectangle.

~~EDIT: Removed this part since I did a mistake, hold on a second!~~
Now I realise that if you were to arrange four rectangles like this:

Then the green area is 4A, and S² is the entire area of the square. So S² - 4A is the area of the white square in the middle - and since you can't have negative area, it must be 0 or more. This makes sense and answers my question.

~~However, I also realise that the difference between the width and height of the rectangles is the same as the sides of the white square. Which is x (and it would make sense that x is real, so that's another, probably more obvious, answer to my previous question). So:~~

x = ± √(S² - 4A)

But using the quadratic formula,

x² - Sx + A = 0

x = (S ± √(S² - 4A))/2

Which is a different answer. Where did I go wrong?
EDIT2: Noticed where I went wrong, I originally labelled the sides of the rectangles in the picture as S - x and S. Instead of S - x and x. So the side of the white square is -S + 2x and not x.

-S + 2x = ± √(S² - 4A)

2x = S ± √(S² - 4A)

x = (S ± √(S² - 4A))/2

Which is the answer I was looking for.