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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
 #1
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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
 #3
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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.

 

-----------------

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.

 

Think about it  laugh

 Aug 16, 2020

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