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 $\sqrt{13}$, what is the value of $a$? Express your answer as a common fraction.
We can solve for the x intersection points thusly
3x + a = x^2 + x
x^2 - 2x - a = 0
x^2 - 2x = a complete the square on x
x^2 - 2x + 1 = a + 1
( x - 1)^2 = a + 1
( x - 1) = sqrt (a + 1) or (x -1) = -sqrt (a + 1)
x= 1 + sqrt (a + 1) or x = 1 - sqrt (a + 1)
The y values of the intersections are
3[1 + sqrt (a + 1)] + a and 3 [ 1 - sqrt (a + 1) ] + a
Simplifying and using the square of the distance formula we have
[ 2sqrt (a + 1) ] ^2 + [ 6sqrt (a + 1)]^2 =13
4 (a + 1) + 36 (a + 1) = 13
40 ( a + 1) = 13
40a + 40 = 13
40a = -27
a = -27 /40
BTW ....the graph looks more like this