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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?

 Jan 24, 2018
 #1
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Set the functions equal

 

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

 

(1/2)x^2  + 3x   =  0

 

Factor

 

x [ (1/2)x + 3 ]  = 0 

 

Setting both factors equal to 0  and solving for x, we have that

 

x   =  0        and   x  =  -6

 

c  = 0     and a  = - 6

 

So   c  - a   =  0 - -6   =  6

 

 

 

cool cool cool

 Jan 24, 2018
edited by CPhill  Jan 24, 2018

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