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

Guest Feb 19, 2018

#1**+2 **

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

ElectricPavlov
Feb 19, 2018

#5**+2 **

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

#7**+2 **

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

#2**+2 **

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

ElectricPavlov
Feb 19, 2018

#8**+1 **

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 :)**

Melody
Feb 21, 2018