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# The parabolas defined by the equations \$y=2x^2-4x+4\$ and \$y=-x^2-2x+4\$ intersect at points \$(a,b)\$ and \$(c,d)\$, where \$c\ge a\$. What is \$c-a

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The parabolas defined by the equations \$y=2x^2-4x+4\$ and \$y=-x^2-2x+4\$ intersect at points \$(a,b)\$ and \$(c,d)\$, where \$c≥a\$. What is \$c-a\$? Express your answer as a common fraction.

Jan 27, 2018

#1
+7563
+3

y  =  2x^2 - 4x + 4    and    y  =  -x^2 - 2x + 4

First let's find the  x  values of the intersection points by solving this equation:

2x^2 - 4x + 4   =   -x^2 - 2x + 4

Add  x^2  to both sides of the equation.

3x^2 - 4x + 4   =   -2x + 4

3x^2 - 2x + 4   =   4

Subtract  4  from both sides.

3x^2 - 2x  =  0

Factor  x  out of the two terms on the left side.

x(3x - 2)   =   0

Set each factor equal to zero.

x  =  0     or     3x - 2  =  0

3x  =  2

x  =  2/3

The  x  coordinates of the intersection points are  0  and  2/3 .

So   c = 2/3   and   a = 0

c - a   =   2/3 - 0   =   2/3

Here's a graph to check it:  https://www.desmos.com/calculator/enndawdskv

Jan 27, 2018

#1
+7563
+3

y  =  2x^2 - 4x + 4    and    y  =  -x^2 - 2x + 4

First let's find the  x  values of the intersection points by solving this equation:

2x^2 - 4x + 4   =   -x^2 - 2x + 4

Add  x^2  to both sides of the equation.

3x^2 - 4x + 4   =   -2x + 4

3x^2 - 2x + 4   =   4

Subtract  4  from both sides.

3x^2 - 2x  =  0

Factor  x  out of the two terms on the left side.

x(3x - 2)   =   0

Set each factor equal to zero.

x  =  0     or     3x - 2  =  0

3x  =  2

x  =  2/3

The  x  coordinates of the intersection points are  0  and  2/3 .

So   c = 2/3   and   a = 0

c - a   =   2/3 - 0   =   2/3

Here's a graph to check it:  https://www.desmos.com/calculator/enndawdskv

hectictar Jan 27, 2018