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The graph of the parabola \(x = 2y^2 - 6y + 3\)  has an x-intercept \((a,0)\)  and two -intercepts \((0,b)\)  and (0,c) . Find a+b+c.

tertre  Mar 12, 2017
 #1
avatar+87334 
+5

x = 2y^2  - 6y + 3

 

When y = 0  , x  = 3   so ( a, 0)   means that a = 3 

 

To find the y intercepts, let x =0    and use the quadratic formula on y

 

 y = [ 6 +/- sqrt (36 - 4(2)(3)] / 4   =  [ 6 +/- sqrt (12)] / 4 =

 

[ 6 +/- 2sqrt(3)]/ 4  =  [3 +/- sqrt(3)]/2

 

So....let b =  [3 + sqrt (3)]/ 2   and let c = [3 - sqrt (3)] / 2

 

So.....a + b + c =

 

3 +   [3 + sqrt(3)]/2  + [3 - sqrt (3)] / 2   =

 

3 +  3/2 + 3/2   =    

 

6

 

Here's a graph that confirms this : https://www.desmos.com/calculator/izhnkrojmb

 

 

 

cool cool cool

CPhill  Mar 12, 2017

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