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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.

 Feb 11, 2020
 #1
avatar+111389 
+1

By Vieta's Theorem

 

The sum of  the roots =  -b / -2

 

So

 

(3 + sqrt 5) + (3 -sqrt 5)  =  -b/-2

 

6  = -b / -2       multiply both sides by  -2

 

-b=  -12

 

b =12

 

 

Also  the product of the  roots  =   c / -2

 

So

 

(3 + sqrt 5) (3 - sqrt 5)  =  c / -2

 

9  - 5  = c / -2

     

4 =  c  /-2     multiply both sides  by -2

 

-8 = c

 

The parabola is

 

y=  -2x^2  + 12x  - 8

 

The x coordinate of the  vertex =  -b /[2a]  =     -12/ [ 2 * -2]  = -12 / -4  =  3

 

And the y coordinate of the  vertex is   -2(3)^2  + 12 (3)  - 8   =    -18 + 36 - 8 =  10

 

So....the vertex  is   ( 3, 10)  

 

 

 

cool cool cool

 Feb 11, 2020
 #2
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+1

Thank you for the help! I never thought of using Vieta's Thereom to solve this. :)

Guest Feb 11, 2020
 #3
avatar+111389 
0

No prob....glad to help   !!!!

 

 

cool cool cool

CPhill  Feb 11, 2020

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