Let b and c be constants such that the quadratic -2x^2+bx+c has roots 3+sqrt5 and 3-sqrt5. Find the vertex of the graph of the equation y=-2x^2+bx+c.

Guest Jan 13, 2019

#1**+1 **

The sum of the roots = -b / a

So....the sum of the roots = 6

So....we have that

-b/ -2 = 6

-b = -12

b = 12

And the product of these roots = c /a

So

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

9 - 5 = c / -2

4 = c / -2

-8 = c

So....the function is y = -2x^2 + 12x - 8

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

So....putting this into the function to find the y coordinate of the vertex we have that

-2(3)^2 + 12(3) -8 = -18 + 36 - 8 = 18 - 8 = 10

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

CPhill Jan 13, 2019