Help me please

There MIGHT (probably) be an easier way, but here is ONE way:

(x+(-3-sqrt5))(x+(-3+sqrt5))   is given in the question for the roots

   if you multiply these two term out you will get

x^2-6x+4    now multiply by -2

-2x^2 +12x-8 =0     (a parabola)

to find vertex take derivative and set equal to zero

-4x+12 = 0     x = 3    substitue this into the equation

        to get      y=10     for the vertex    3,10

OK.....another way...

AFTER you get to

-2x^2 +12x -8                arrange into vertex form  y=a(x-h)^2 +k    the vertex is h,k

-2{ x^2 -6x +4}

-2{(x-3)^2 +4 -9}

-2 {(x-3)^2 -5}

-2 (x-3)^2 +10             so  h,k is the vertex   =   3,10

I am quite sure EP will be correct but I will take a look:

Let b and c be constants such that the quadratic $-2x^2 +bx +c$   

has roots $3+\sqrt{5}$  and $3-\sqrt{5}$.

Find the vertex of the graph of the equation   

$y=-2x^2 + bx + c$

The roots of a quadratic   $ax^2+bx+c$  is the answers to   $ax^2+bx+c=0$

And the answers to this are given by the quadratic equation

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=3\pm\sqrt5$

This means that  $x=\frac{-b}{2a}$ = 3      must be exactly halfway between the two roots.!!

So the axis of symmetry is x=3 and the vertex lies on this line so the x value of the vertex is 3

The y value will be    $y=-2*3^2+bx+c = -18+3b+c$

This is fine but you need to find the vleu of b and c       

From the equation above I can see that 

$\frac{-b}{2a}=3 \qquad and \qquad \pm\frac{\sqrt{b^2-4ac}}{2a}=\pm\sqrt5\\ a=-2\qquad so\\ \frac{-b}{-4}=3 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{-4}=\pm\sqrt5\\ b=12 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{4}=\pm\sqrt5\\ b=12 \qquad and \qquad \frac{b^2+8c}{16}=5\\ b=12 \qquad and \qquad b^2+8c=80\\ b=12 \qquad and \qquad 144+8c=80\\ b=12 \qquad and \qquad 18+c=10\\ b=12 \qquad and \qquad c=-8\\$

so

$y= -18+3b+c\\ y= -18+3*12+-8\\ y=-18+36-8\\ y=10$

So the vertex is   (3,10)

Which is exactly the same answer has given you. He has done it a number of different ways but stlill all the answers are the same!!

Thanks EP :)