Probability of Root Placement

1) For roots to be real $k^2-4 \geq 0$ which gives $k \geq 2$ or $k \leq -2$

2) Value of $f(x)$ should be positive at $x=-2$ and $x=4$ which gives conditions $k \leq 2.5$ & $k \geq -4.25$ respectively.

3) Vertex of parabola must be between -2 and 4, which gives $4 > k > -8$ .

From the above conditions $k$ belongs to $[-4.25,-2]\cup [2,2.5]$ . Hence, Length of $k =2.75$ , which gives $p = \frac {2.75}{10}$ , $\lfloor 1000p \rfloor =275$ .

[Latex edits - Calvin]

We note that the given function always has a positive leading coefficient, so the quadratic is a parabola that opens upwards. As such, for the quadratic to have roots between -2 and 4, it is necessary and sufficient to satisfy the following 4 conditions:

1. It must have real roots. Hence $\mbox{Discriminant} \geq 0$ $\Rightarrow k^2 -4 \geq 0 \Rightarrow k \leq -2$ or $k \geq 2$ .

2. $f(-2) \geq 0$ . Hence $\Rightarrow 4 - 2k + 1 \geq 0 \Rightarrow k \leq \frac{5}{2}$

3. $f(4) \geq 0$ . Hence $\Rightarrow 16 + 4k + 1 \geq 0 \Rightarrow k \geq \frac{-17}{4}$

4. The vertex of the parabola must lie between -2 and 4. Hence $-2 \leq \frac {-k} {2} \leq 4 \Rightarrow -8 \leq k \leq 4$ .

Putting these together we have two possible intervals for $k$ : $- \frac {17}{4} \leq k \leq -2$ and $2 \leq k \leq \frac {5}{2}$ .

Hence, the probability is $p = \frac{\left(-2 - \left(-\frac{17}{4}\right) \right) + \left(\frac{5}{2} - 2\right)}{5 - (-5) } = \frac {\frac {11}{4}}{10} = .275$ . Thus $\lfloor 1000p \rfloor = 275$ .

The roots of $f(x)$ by the quadratic formula is $\frac{-k\pm \sqrt{k^2-4}}{2}$ . So, we must have $-2\leq \frac{-k-\sqrt{k^2-4}}{2}\leq 4 \Rightarrow 2\leq k\leq 2.5$ or $k\leq -2$ and $-2\leq \frac{-k+ \sqrt{k^2-4}}{2}\leq 4\Rightarrow 2\leq k$ or $-4.25\leq k\leq -2$ . Combining these 2 inequalities, we get $-4.75\leq k \leq -2$ or $2\leq k\leq 2.5$ . Then $p=\frac{(2.5-2)+(-2-(-4.25))}{5-(-5)}=\frac{2.75}{10}$ . Therefore, $\lfloor 1000p\rfloor=275$ .

The two roots of $f(x) = x^2 + kx + 1$ are $\frac{-k \pm \sqrt{k^2-4}}{2}$ . We need these to be real, so $k \ge 2$ or $k \le -2$ . If they are real, then they must be between $-2$ and $4$ , so $-2 \le \frac{-k \pm \sqrt{k^2-4} }{2} \le 4$ . Rearranging, we get that $-4+k \le \pm \sqrt{k^2-4} \le k+8$ . We only need to consider $k-4 \le -\sqrt{k^2-4}$ and $\sqrt{k^2-4} \le k+8$ , because the other cases are true if these two are true. Solving the first inequality, we get that $k \le \frac 52$ , and solving the second inequality, we get that $k \ge -\frac{17}{4}$ . Thus $x \in [-\frac{17}{4},-2] \cup [2,\frac{5}{2}]$ , so $p = \frac{11}{40}$ , so $\left \lfloor 1000p \right \rfloor = 275$ .

p=0.275

From $x^{2} + kx + 1 = 0$ , we get $x = \frac{-k \pm \sqrt{k^{2} - 4}}{2}$ . From the roots are between $-2$ and $4$ , we conclude that $-4 < -k \pm \sqrt{k^{2} - 4} < 8$ , and therefore $\sqrt{k^{2} - 4} > k - 4$ , $\sqrt{k^{2} - 4} > -(k + 8)$ , $\sqrt{k^{2} - 4} < 4 - k$ and $\sqrt{k^{2} - 4} < k + 8$ .

Now, consider $4$ cases.

Case $1$ : $-5 \le k \le -2$

Then $k - 4 \le -(k + 8), k + 8 \le 4 - k$ . Thus, we just need to consider $\sqrt{k^{2} - 4} > -(k + 8), \sqrt{k^{2} - 4} < k + 8$ . The first inequality clearly holds since in $[-5, 5]$ , $-(k + 8) < 0 \le \sqrt{k^{2} - 4}$ , and the second holds if and only if $- \frac{17}{4} \le k \le -2$ . Since both sides of the second inequality is nonnegative, we can square both sides and conclude that $k \ge - \frac{17}{4}$ , and thus the conclusion that if $-5 \le k \le -2$ , then $- \frac{17}{4} \le k \le -2$ .

Case $2$ : $-2 < k < 2$

Then the discriminant, $k^{2} - 4 = (k + 2)(k - 2) < 0 \Rightarrow x^{2} + kx + 1 = 0$ has no real roots, which is a contradiction.

Case $3$ : $2 \le k < 4$

Since $k - 4 > -(k + 8), k + 8 > 4 - k$ , and so we just need to consider $k - 4 < \sqrt{k^{2} - 4} < 4 - k$ . The LHS follows from $\sqrt{k^{2} - 4} \ge 0 > k - 4$ , since $k < 4$ . And the RHS holds if and only if $2 \le k < \frac{5}{2}$ , since by $4 - k, \sqrt{k^{2} - 4} \ge 0$ and so we can square both sides to conclude $2 \le k < \frac{5}{2}$ ( $k \ge 2$ ) is from the assumption.

Case $4$ : $4 \le k \le 5$

Since $k - 4 > -(k + 8), k + 8 > 4 - k$ , and so we again just need to consider $k - 4 < \sqrt{k^{2} - 4} < 4 - k$ . But this time the RHS doesn't hold since $\sqrt{k^{2} - 4} \ge 0 \ge 4 - k$ , by $k \ge 4$ , which is a contradiction and thus we have no solution.

In conclusion, the solutions are $- \frac{17}{4} \le k \le -2$ and $2 \le k < \frac{5}{2}$ .

Now, it is now trivial to compute that $\lfloor 1000p \rfloor = \boxed {275}$ .

the roots of the equation X^2+kX+1=0 is (-k+(k^2-4)^.5)/2 & (-k-(k^2-4)^.5)/2 the roots are between -2 and 4 . there fore we have k-4 less equal (k^2-4)^.5 less equal k+8 . and k-4 less equal -(k^2-4)^.5 less equal k+8 . now we can easily find out the solution of k for the real roots . the solution is -4.25 less equal k less equal -2 and 2 less equal k less equal 2.5 . hence the total length of interval where we can find k is (4.25-2) + (2.5-2) =2.75 the function f(x) defined in the closed interval -5 to 5 . the length of the interval is (5+5)=10. therefore the required probability is 2.75/10 ie 275/1000. hence the answer is (275/1000)*1000 = 275 .

=

f(x)=x2+kx+1. roots of this equation are (-k+/-√(k^2-4))/2. For least k value the root must be less than 4 thus (-k+/-√(k^2-4))/2=4 gives least k, similarly (-k+/-√(k^2-4))/2=-2 gives highest k ,solving those equations for k we get least value of k as -4.25 and highest value of k as 2.5. If the k value is between -2 and 2 then roots becomes imaginary ( k^2<4). So the actual k values are -4.25 to -2 and 2 to 2.5 .Thus constituting 2.75 (-2-(-4.25)+2.5-2=2.75). The overall interval is [−5,5] ,constituting 10(5-(-5)). Thus p=(2.75)/10 and ⌊1000p⌋=275

If we look at the problem from a graphical and calculus point of view........ The conditions just boils down to the quadratic function having a minima at x say m for which m lies between (-2,4) And also the vertex of the parabola to lie in the lower side of the coordinate plane . now this condition is given by f(m) < 0. Now f'(x) =0 for x= -k/2. Or m=-k/2 So the first condition is given by. f(-k/2) <0 . And also the sign of f(-2) and f(4) must be the same(in this case positive). So we get our second condition. Plug these together to find that k lies in (-4.25, -2) U (2, 2.5). Giving us the probability as 275/1000.

I think the answer may be slightly flawed. Correct me if I'm wrong on this because I'm curious. Since the question asks for the floor function of 1000p, I think the probability should drop to 274. But, then again, since we are essentially looking at an infinite / continuous range of values, .999999999999999999 repeating =1 so rip. ^Somebody explain this better lol/. I'm mixed with opinions. Help

Let Z be the roots of the quadratic f(x).

$\frac {-k \pm \sqrt{k^2 - 4}}{2}$

We are asked to find P[-2 < Z < 4].

$P[-2 < Z < 4] = P[-2 < \frac {-k \pm \sqrt{k^2 - 4}}{2} < 4]$ for -2<Z<-1 or 1<Z<4

$P[-2 < Z < 4] = P[-2 < \frac {-k - \sqrt{k^2 - 4}}{2} < -1] + P[1 < \frac {-k + \sqrt{k^2 - 4}}{2} < 4]$

We can solve K by solving the left side and the right side of the inequality separately.

ie. $\frac{-k - \sqrt{k^2 - 4}}{2} < -1$

$k + \sqrt{k^2 - 4} > 2$

$k^2 - 4 > k^2 - 4k + 4$

$k > 2$

After applying to all sides, we arrive at the following:

$P[-2 < Z < 4] = P[2 < K < 2.5] + P[-4.25 < K < -2]$

The probability density function of the continuous uniform distribution is: $f(x) = \left\{ \begin{array}{ll} \frac{1}{b-a}\ & \quad a \leq x \leq b \\ 0 & \quad x < a\ or\ x > b \end{array} \right.$

$P[2 < K < 2.5] = \int_2^{2.5} \frac{1}{5 - (-5)}\,\mathrm{d}x = 0.05$

$P[-4.25 < K < -2] = \int_{-4.25}^{-2} \frac{1}{5 - (-5)}\,\mathrm{d}x = 0.225$

$P[-2 < Z < 4] = 0.05 + 0.225 = 0.275$

$\lfloor{1000*0.275}\rfloor = 275$

First, we note that the roots are: $\frac{-k\pm\sqrt{k^2-4}}{2}$ . Since it must have two roots, $k^2-4\geq 0$ , therefore $-2 < k < 2$ .

Next we note also that $-2\leq\frac{-k\pm\sqrt{k^2-4}}{2}\leq 4$ or $-4\leq -k\pm\sqrt{k^2-4}\leq 8$ . Since $\sqrt{k^2-4}$ is positive, it is equivalent to saying $-4\leq -k-\sqrt{k^2-4}$ and $-k+\sqrt{k^2-4}\leq 8$ .

For $-4\leq -k-\sqrt{k^2-4}$ , which is equivalent to $\sqrt{k^2-4}\leq 4-k$ , since $\sqrt{k^2-4}$ is positive, we can square both sides to get the final inequality $k\leq 2.5$ .

For $-k+\sqrt{k^2-4}\leq 8$ , which is equivalent to $\sqrt{k^2-4}\leq 8-k$ , since $\sqrt{k^2-4}$ is positive, we can square both sides to get the final inequality $-4.25\leq k$ .

Combining the above three inequality, we have: $-4.25\leq k < -2$ or $2 < k \leq 2.5$ , giving us the total interval length of $2.75$ . So the probability is $p=\frac{2.75}{10}=0.275$ , therefore $\lfloor 1000p\rfloor = 275$

$x = \frac{-k \pm \sqrt{k^2-4}}{2}$ so we can find the maximum and minimum of k along the interval [-5,5]

$4 = \frac{-k + \sqrt{k^2-4}}{2} \Rightarrow k = \frac{5}{2}$

$-2 = \frac{-k - \sqrt{k^2-4}}{2} \Rightarrow k = -\frac{17}{4}$

But to keep the roots real, $\sqrt{k^2-4} \geq 0 \Rightarrow k \geq 2$ or $k \leq -2$

Thus the valid values is $-\frac{17}{4} \leq k \leq -2$ and $2 \leq k \leq \frac{5}{2}$

$-2+\frac{17}{4} + \frac{5}{2}-2 = \frac{11}{4} \Rightarrow \frac{11}{4}*100 = 275$

And the floor of that is still $275$

Determinant=k^2-4 \geq 0 Thus k \geq 2 or k \leq -2. As both roots are in between -2 and 4, -2 \leq \frac{-k \pm \sqrt{k^2-4}}{2} \leq 4. i.e. k-4 \leq \pm \sqrt{k^2-4} \leq k+8 Now we consider the 2 cases k \geq 2 and k \leq -2 separately. Case 1: If 2 \leq k \leq 5, then \pm \sqrt{k^2-4} \leq k+8 is always true for all k as the minimum value of k+8 (which is 10) is greater than the largest value of \pm \sqrt{k^2-4}. Similarly k-4 \leq \sqrt{k^2-4} is always true for all k in the interval of [2,5]. The only sub-case left is k-4 \leq -\sqrt{k^2-4}, which is 4-k \geq \sqrt{k^2-4} in other words. Notice that k /leq 4 in this case. Solving this inequality by taking square of both sides, we obtain the following solution 2 \leq k \leq 2.5. Case 2: If -5 \leq k \leq -2, similarly we can conclude that both k-4 \leq \pm \sqrt{k^2-4} and -\sqrt{k^2-4} \leq k+8 are always true for all k in the interval of [-5,-2]. The only sub-case left is \sqrt{k^2-4} \leq k+8. This can be solved by taking the square of both sides and the following solution is obtained: -4.25 \leq k \leq -2. Therefore p= \frac {(2.5-2) + (-2-(-4.25))}{5-(-5)}=0.275 Hence, 1000p=275

The roots of $f(x)$ is $\frac{-k\pm \sqrt{k^2-4}}{2}$ . So, we must have $-2\leq \frac{-k+\sqrt{k^2-4}}{2}$ and $4\geq \frac{-k- \sqrt{k^2-4}}{2}$ ,then we will have $k\leq 2.5$ and $k\geq -4.25$ . But we must have $k^2\geq 4$ to obtain real roots. So, $k\geq 2$ and $k\leq -2$ . The probability is $\frac{(2.5-2)+[-2-(-4.25)]}{5-(-5)}=\frac{2.75}{10}$ . Therefore, $\lfloor 1000p\rfloor=\lfloor1000\cdot \frac{2.75}{10}\rfloor =275$ .Done.