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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.

 Jan 13, 2019
 #1
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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)

 

 

cool cool cool

 Jan 13, 2019
edited by CPhill  Jan 13, 2019
edited by CPhill  Jan 13, 2019
 #2
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Thank you so much!!!

Guest Jan 13, 2019

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