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A quadratic of the form \(-2x^2+bx+c\) has roots of \(x=3+\sqrt5\) and \(x = 3-\sqrt5\). The graph of \(y​​​​=\) \(-2x^2+bx+c\) is a parabola. Find the vertex of this parabola.

 Jul 5, 2020
 #1
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Here is the quadratic equation

 

(x -3 - sqrt5)(x-3+sqrt5) = 0       Multiplying it out

 

x^2    -3x +sqrt5 x          -3x +9 -3sqrt5            - sqrt5 x   +3 sqrt 5   -   5

x^2    -6x    +4   = 0         to get it into the form asked about in the questiion , multiply through by -2

-2x^2  +12x  -8    = 0 

 

 Now find 'x' coordinate of the vertex by    - b /2a          b = 12    a = -2

 

Use THIS 'x' value in the final equation to find the 'y' coordinate of the vertex....

 Jul 5, 2020
 #2
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Thank you! Could you please explain farther?

 Jul 6, 2020

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