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# A quadratic of the form has roots of and . The graph of is a parabola. Find the vertex of this parabola.

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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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(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