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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

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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 Jan 24, 2018
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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

CPhill  Jan 24, 2018
edited by CPhill  Jan 24, 2018