A quadratic of the form -2x^2+bx+c has roots of x=3+sqrt(5) and x=3-sqrt(5). The graph of y=-2x^2+bx+c is a parabola. Find the vertex of this parabola.

Guest Feb 11, 2020

#1**+1 **

By Vieta's Theorem

The sum of the roots = -b / -2

So

(3 + sqrt 5) + (3 -sqrt 5) = -b/-2

6 = -b / -2 multiply both sides by -2

-b= -12

b =12

Also the product of the roots = c / -2

So

(3 + sqrt 5) (3 - sqrt 5) = c / -2

9 - 5 = c / -2

4 = c /-2 multiply both sides by -2

-8 = c

The parabola is

y= -2x^2 + 12x - 8

The x coordinate of the vertex = -b /[2a] = -12/ [ 2 * -2] = -12 / -4 = 3

And the y coordinate of the vertex is -2(3)^2 + 12 (3) - 8 = -18 + 36 - 8 = 10

So....the vertex is ( 3, 10)

CPhill Feb 11, 2020