# Does the parabola y = 4x^2 - 4x + 8 intersect the line y = 10

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### 3 Answers

If two curves intersect each other, the coordinates of the point at which they do so have to satisfy the equations that define both the curves. To determine if the parabola y = 4x^2 - 4x + 8 intersects the line y = 10, solve the equation 10 = 4x^2 - 4x + 8. If the roots of the equation are real, the two intersect. If the roots of the equation are complex the parabola does not intersect the line. There could also be third possibility, the line is a tangent of the parabola if the roots of the equation are equal.

4x^2 - 4x + 8 = 10

=> 4x^2 - 4x - 2 = 0

=> 2x^2 - 2x - 1 = 0

The roots are `(2+-sqrt(4+4*2))/4 = (1+-sqrt3)/2`

As the roots of the equation are real and different, the parabola y = 4x^2 - 4x + 8 intersects the line y=10.

To determine the points of intersection of the parabola y = 4x^2 - 4x + 8 and the line y = 10, solve the equation 10 = 4x^2 - 4x + 8. The solution of the equation is the x-coordinates of the line with the y-coordinate for both being the same and equal to 10.

10 = 4x^2 - 4x + 8

4x^2 - 4x - 2 = 0

2x^2 - 2x - 1 = 0

`(sqrt 2*x)^2 - 2*sqrt 2*(1/sqrt 2)*x + 1/2 - 3/2 = 0`

`(sqrt 2*x - 1/sqrt 2)^2 = 3/2`

`sqrt 2*x - 1/sqrt 2 = +-sqrt 3/sqrt 2`

`2*x - 1 = +-sqrt 3`

`x = (1 +- sqrt 3)/2`

The points of intersection are `((1 - sqrt 3)/2, 10)` and `((1 + sqrt 3)/2, 10)`

Just by substituting the solutions,we can find the line intersects