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\(x^2 + 4y^2 = 4x\\ (x-2)^2 + 4y^2 = 4\\ y^2 = 1 - \dfrac{(x-2)^2}{4} ,~x \in [0,4]\\ x^2 + y^2 = x^2 + 1 - \dfrac{(x-2)^2}{4} = \\ \dfrac 3 4 \left(x+\dfrac 2 3\right)^2 - \dfrac 1 3\)

\(\text{This is clearly maximum at the maximum value of }x \text{ i.e. }x=4\\ \dfrac 3 4 \left(4 + \dfrac 2 3\right)^2 - \dfrac 1 3 = 16\)

\(x+y = x + \sqrt{1 - \dfrac{(x-2)^2}{4}}\\ \text{we're going to have to use calculus to maximize this}\\ \dfrac{d}{dx} \left( x + \sqrt{1 - \dfrac{(x-2)^2}{4}}\right) =\\ \dfrac{1}{2} \left(\dfrac{2-x}{\sqrt{(4-x) x}}+2\right) = 0 \Rightarrow \\ x = \dfrac{2}{5} \left(5+2 \sqrt{5}\right),~y=\dfrac{\sqrt{5}}{5} \\ x+y = 2+\sqrt{5}\)

Maybe you can follow this.

-16x^4 + x^2 + 2x + 1       =

x^2 + 2x + 1 - 16x^4    =

(x + 1)^2 - 16x^4   =

[ (x + 1) + 4x^2] [ (x + 1) - 4x^2 ] =

[  4x^2 + x + 1 ]    [ -4x^2 + x + 1 ]

And   a > d

Thanks Rom :)

Another interesting one for me to inspect :))

Thanks Rom,

I will have to think about that one :)

How about saying thank you Mathtoo ...    

I'm fine, just counting down to new year :D

Wow tertre, that's REALLY cool! :D

it's pretty clear P(x) can't be an odd degree so we'll try a 2nd degree polynomial

\(\text{Letting }P(x) = c_0 + c_1 x + c_2 x^2\\ \text{then grinding out the expansions of each side and equating like power of x coefficients}\\ \text{we get two solutions}\\ P(x) = 0\\ P(x) = x+x^2\)

That is correct! Now, try to solve it when a>d.

Factor the following:
-16 x^4 + x^2 + 2 x + 1

Factor -1 out of -16 x^4 + x^2 + 2 x + 1:
-(16 x^4 - x^2 - 2 x - 1)

The factors of 1 - 16 x^4 that sum to -2 are x (-4 x^2 - 1) and x (4 x^2 - 1). 
So, (16 x^4 - 1) - x^2 - 2 x = -x^2 + x (-4 x^2 - 1) + x (4 x^2 - 1) + (4 x^2 - 1) (4 x^2 + 1):
-(-x^2 + x (-4 x^2 - 1) + x (4 x^2 - 1) + (4 x^2 - 1) (4 x^2 + 1))

-x^2 + x (-4 x^2 - 1) + x (4 x^2 - 1) + (4 x^2 - 1) (4 x^2 + 1) = x ((4 x^2 - 1) - x) + (4 x^2 + 1) ((4 x^2 - 1) - x):
-x (4 x^2 - x - 1) + (4 x^2 + 1) (4 x^2 - x - 1)

Factor 4 x^2 - x - 1 out of x (4 x^2 - x - 1) + (4 x^2 + 1) (4 x^2 - x - 1), resulting in (4 x^2 - x - 1) ((4 x^2 + 1) + x):

 -(4 x^2 - x - 1) (4 x^2 + x + 1)

Okay, thanks so much, CPhill!

I've only estimated graphically

Looks like about 6.2

I do not know but it looks interesting. I will be interested in the answer when it comes  

Well noobie, you are off to a good start   

Since the decimal terminates  ⇒   rational

(x - 1/A) (x - 1/B) (x - 1/C) = 0

(AX - 1) (BX - 1) (Cx - 1) / (ABC) = 0

(Ax - 1) (Bx - 1) (Cx - 1) = 0

[ (ABx^2 - (A + B)x  + 1 ] [ Cx - 1 ] = 0

(ABC)x^3 - (AC + BC)x^2 + Cx - ABx^2 + (A + B)x - 1 = 0

(ABC)x^3 - (AC + BC + AB)x^2 +(A + B + C)x  - 1 = 0

Write the original polynomial as    x^3 + 0x^2 - 5x - 1  =  ax^3 + bx^2 +cx + d

a = 1    b = 0    c = -5     d = - 1

Adding the original roots in the original polynomial = (A + B + C) =   -b/a  = -0/ 1   = 0

Multiplying the roots gives (ABC) =  -d /a =   - (-1) /1  = 1

And (AC + BC + AB) =  c/a    = -5 /1 = - 5

So

The new polynomial is

(1) x^3  - (- 5)x^2   - 1   =

x^3 + 5x^2 - 1

That's what I had originally said, but it said that I was wrong... confusing...

I think that (2) is the true statement.......

The circumcenter is where the perpedicular bisectors meet

The orthocenter is where the altiudes intersect

In an equilateral triangle, these are the same from any vertex

Thanks, I appreciate it! Though I'm so glad that I was able to find this website!! So I should be the one who's thanking you!!

No prob....every now and then I manage to get one correct....LOL!!!

BTW....welcome aboard  !!!

Oh, my bad. I didn't even write the question properly... just edited though, so it should be good now!