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

 Feb 19, 2018

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

 Feb 19, 2018
edited by ElectricPavlov  Feb 19, 2018

Should it be (x+(3-sqrt 5))(x+(3+sqrt5))?

Guest Feb 21, 2018

Shouldnt it be (-3,10)?

Guest Feb 21, 2018

The roots are given   as   3+-sqrt5

so one part would be


x+(- 3+sqrt5) =0  

x-3+sqrt5  = 0     solve for x   (add  (3 and subtract sqrt5) from both sides )

x= 3 -sqrt5



   and the other would be  x+( -3-sqrt5)  =0

            x-3-sqrt5    solve for x  (add 3 and sqrt 5 to both sides  

                                          x = 3+sqrt5


These are the roots given in the question....

ElectricPavlov  Feb 21, 2018

Here is a graph  

ElectricPavlov  Feb 21, 2018

y=  a (x-h)^2  +k        and

y= -2 (x-3)^2 +10


Now can you see that a =-2    h=3   k = 10  ???

ElectricPavlov  Feb 21, 2018

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

 Feb 19, 2018

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       frown


From the equation above I can see that 

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




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


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 :)

 Feb 21, 2018

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