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

Guest Feb 19, 2018
#1
+12560
+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
edited by ElectricPavlov  Feb 19, 2018
#3
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Should it be (x+(3-sqrt 5))(x+(3+sqrt5))?

Guest Feb 21, 2018
#4
0

Shouldnt it be (-3,10)?

Guest Feb 21, 2018
#5
+12560
+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
#6
+12560
+2

Here is a graph

ElectricPavlov  Feb 21, 2018
#7
+12560
+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
+12560
+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
+92805
+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   $$ax^2+bx+c=0$$

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

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