Is this the problem?

Algebra Level 5

Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\left \lfloor\sqrt{P(x)}\right \rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\dfrac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a, b, c, d,$ and $e$ are positive integers. Find $a + b + c + d + e$ .

The answer is 850.

1 solution

Siddharth Iyer
Apr 15, 2015

We note that P(x) is strictly increasing for x > 1.5. Now integer(x) is always an integer and it is an integer in the interval [5,15]. We check integers from 5 to 15 to see whether the RHS is an integer. We see that this is true if and only if integer(x)=5,6 and 13. Corresponding values of RHS being 1,3 and 11. Now we consider intervals around x=5,6 and 13. For all x where (P(x)^0.5) is in [1,2), [3,4) and [11.12) where x is in [5,6), [6,6) and [13,14) respectively. Now we only need to solve for the upper bound of P(x) in each set as we know that P(x) is strictly increasing. Solving gives x is in [5, 0.5(3+(61)^0.5)), [6, 0.5(3+(109)^0.5)) and [13, 1.5(1+(69)^0.5)). Now we can compute the total length of the intervals and divide by 15 -5 = 10. To give the required solution.

Is this the problem?

The answer is 850.

1 solution

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