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.

Guest Aug 14, 2020

#1**+1 **

(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

Guest Aug 14, 2020

#3**+2 **

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.

Think about it

Melody Aug 16, 2020