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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 $$x = 3 + \sqrt{5}$$ and $$x = 3 - \sqrt{5}.$$The graph of $$y = -2x^2 + bx + c$$ is a parabola. Find the vertex of this parabola.

Aug 14, 2020

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(x-3-sqrt5)(x-3+sqrt5)

Multiply it out.....    then multiply by -2 to get it in the required form

then you will have  a  b   and  c      vertex will be at   x = - b/2a

use this value of 'x' in the quadratic to calculate the 'y' component

Aug 14, 2020
#2
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Why are you multiplying the roots toghter?

Guest Aug 15, 2020
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Try thinking about a different question.

find the roots of    $$y=x^2+5x+6$$

To do this (without a formula) I must first factorize it.

$$y=(x+2)(x+3)$$

Now if x=-2 then y will be 0

and if x=-3 then y will also be 0.

So

x=-3 and x=-2 are roots of this equation.

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Now I will go in the other direction:

If  x=-3 and x=-2 are roots of     $$y=x^2+5x+c$$     find c.

I would say

y=(x--3)(x--2)

y=(x+3)(x+2)

$$y=x^2+5x+6$$

so     c=6.