+266 The graph of the line y = 3x+a intersects the graph of the parabola y = x^2+x in two points. If the distance between these points is 3\(\sqrt{30}\), what is the value of a? Express your answer as a common fraction.
Please include the solution instead of just the answer! That would be much appreciated. Also, if any moderators would help, that would be great
To find the value of a, we need to determine the two points of intersection between the two functions. Let's call the two points of intersection (x1,y1) and (x2,y2).
Setting y = 3x + a equal to y = x^2 + x, we get:
3x + a = x^2 + x
3x + a - x^2 - x = 0
x^2 + 2x + (3-a) = 0
We can now use the quadratic formula to find the solutions for x:
x = (-b ± √(b^2 - 4ac)) / 2a
x = (-2 ± √(2^2 - 4 * 1 * (3-a))) / 2 * 1
x = (-2 ± √(4 + 4a - 4 * 3)) / 2
x = (-2 ± √(4a - 8)) / 2
Since the two points of intersection are distinct, the square root must be real and nonnegative, so:
4a - 8 ≥ 0
a ≥ 2
The two points of intersection can now be found by substituting the values of x back into y = 3x + a:
y1 = 3x1 + a = 3 * (-2 + √(4a - 8)) / 2 + a = -3 + 3√(4a - 8) + a
y2 = 3x2 + a = 3 * (-2 - √(4a - 8)) / 2 + a = -3 - 3√(4a - 8) + a
Finally, the distance between the two points is given as 3 * sqrt(15), so:
√((x1 - x2)^2 + (y1 - y2)^2) = 3 * √(30)
√((-2 + 2√(4a - 8))^2 + (3 + 3√(4a - 8) - (-3 - 3√(4a - 8)))^2) = 3 * √(30)
√(4 + 6√(4a - 8)) = 3 * √(30)
4 + 6√(4a - 8) = 15
6√(4a - 8) = 11
√(4a - 8) = 11 / 6
√(4a - 8) = √(11 / 6)
4a - 8 = 11 / 6 * 11 / 6
4a = 11 / 6 * 11 / 6 + 8
4a = 121 / 36 + 144 / 36
4a = 265 / 36
a = 265 / 36 * 9 / 4
a = 765 / 144
So the value of a is 765/144. This is already in the form of a common fraction, so this is our final answer.
