The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two points P and Q. The length of PQ is 410. What is

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Assume the intersections are P(p, 3p+b) and Q(q, 3q+b).
Then the length of PQ is
\(\sqrt{\left(p-q\right)^2+\left(3p+b-3q-b\right)^2}\)
\(=\sqrt{\left(p-q\right)^2+\left(3p-3q\right)^2}\)
\(=\sqrt{\left(p-q\right)^2+9\left(p-q\right)^2}\)
\(=\sqrt{10\left(p-q\right)^2}\)
\(=\sqrt{10\left[\left(p+q\right)^2-4pq\right]}\)
\(=4\sqrt{10}=\sqrt{160}\)

Then we have 10[(p+q)^2-4pq]=160 or (p + q)^2 - 4pq = 16.
Since p and q are roots of the equation x^2 + ax + 4 = 3x + b or x^2 + (a - 3)x + 4 - b = 0, we have (x - p)(x - q) = x^2 - (p + q)x + pq = x^2 + (a - 3)x + 4 - b, -(p + q) = a – 3, or p + q = -a + 3 and pq = 4 - b.
Then (-a + 3)^2 - 4(4 - b) = 16, (-a + 3)^2 - 16 + 4b = 16, or 4b = -(-a + 3)^2 + 32.
We have b = -(1/4)(-a + 3)^2 + 8 and the maximum value of b = 8 at a = 3.

Therefore, D is the answer.
Answer: D