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The parabolas defined by the equations $y=x^2+4x+6$ and $y=\frac{1}{2}x^2+x+6$

intersect at points $(a,b)$ and $(c,d)$, where $c\ge a$. What is $c-a$?

Guest Mar 31, 2017
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\( $y=x^2+4x+6$ and $y=\frac{1}{2}x^2+x+6$\)     \($c\ge a$\)

 

Set these equal

 

x^2 + 4x + 6   = (1/2)x^2 + x + 6  ...... rearrange as

 

(1/2)x^2 + 3x  = 0   multiply through by 2

 

x^2 + 6x  = 0      factor

 

x ( x + 6)  = 0      set both factors to 0  and solve for x and x = 0  or x  = -6

 

And when x  = 0,   y  =  (0)^2 + 4(0) + 6   = 6

And when x = - 6, y = (-6)^2 + 4(-6) +  6  = 18

 

So...... the intersection points are  (0, 6)  and ( -6 , 18)

 

And c - a  =  0 - (-6)    =   6

 

Here's a graph : https://www.desmos.com/calculator/txramx7pv5

 

 

cool cool cool

CPhill  Mar 31, 2017

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