Probability of Distinct Roots

$f(x) = kx^4 + (k^2+1)x^2 + k = k(x^2+k)(x^2 + \frac {1}{k})$ . Hence the roots are $x = \sqrt{-k}, -\sqrt{-k}, \sqrt{-\frac {1}{k}}, -\sqrt{-\frac {1}{k}}$ . Thus, for $f(x)$ to have 4 real roots, $k$ must be negative. The roots $\sqrt{-k}$ and $\sqrt{ -\frac {1}{k}}$ are strictly positive, and clearly greater than -1. Hence, one of the negative roots is less than -4, and the other is more than -1.

Case 1: $-\sqrt{-k} < -4$ . This implies that $k < -16$ which is out of the domain.

Case 2: $- \sqrt{-\frac {1}{k} } < -4$ . This implies that $0> k> -\frac {1}{16}$ . Furthermore, we will have $- \sqrt{-k} > -\frac {1}{4} > -1$ , hence the second root is more than -1.

As such, the possible range of $k$ is $( - \frac {1}{16}, 0)$ . Hence $p = \frac { 0-(-\frac {1}{16} )} { 5 - (-5) }= 0.00625$ . Therefore, $\lfloor 1000p \rfloor = 6$ .

The solution of f(x) = 0 can be written as:

x= /pm /sqrt{\frac{-(k^2+1) \pm \sqrt{(k^2+1)^2-4k^2}}{2k}}

Fortunately, after expansion and simplification, this expression becomes much more manageable. For the four possible combinations of the /pm of the two square roots, (++, +-, -+, and --) we have:

1. (++) --> x = /sqrt{/frac{-1}{k}}

2. (+-) --> x = +sqrt{-k}

3. (-+) --> x = -sqrt{/frac{-1}{k}}

4. (--) --> x = -sqrt{-k}

From these equation it is clear that x must be less than 0 to obtain all real roots. In addition, we see that equation 3 is the only equation that can produce a number less than -4 for -5<k<5 (by choosing a value of k negative and close to 0.) The exact interval for which a number less than -4 is produced can be calculated:

-4 = -sqrt{/frac{-1}{k}}

16*k = -1

k = /frac{-1}{16}

So any value of k in between -1/16 and 0 gives x<-4. This is the key realization, because we can easily see that any value of k in this interval also gives x-values greater than -1 in equations 1, 2, and 4.

Hence, -1/16<k<0 is the largest interval in which these conditions are satisfied.

The rest is just a simple calculation: We have, for the probability p that the conditions are fulfilled:

p = /frac{1/16}{10} = /frac{1}{160}

So,

1000p = 6.25

The floor of 6.25 is 6--This is the answer.

Observe that the roots of $f$ come in pairs $x, -x$ . Hence, $f$ has a root that is less than -4, a root that is between -1 and 0, a root that is between 0 and 1, and a root that is above 4. Use the substitution $x^2=t$ , and consider the quadratic $g(t) = kt^2 + (k^2+1)t + k$ . Then $g(t)$ has a root that is between 0 and 1, and a root that is above 16.

We consider the following cases:

Case 1: $k>0.$ In this case, the quadratic g(t) opens up, hence the condition on the roots of $g(t)$ imply that $g (0) > 0, g(1)<0$ and $g(16)<0$ . As such, $k$ must satisfy $k>0, k+ (k^2+1)+k <0$ and $256k + (16k^2+1) + k < 9$ . Note that the second inequality is $(k+1)^2 < 0$ which is never true. Thus, $k>0$ is not possible.

Case 2: $k < 0.$ In this case, the quadratic $g(t)$ opens down, hence the condition on the roots of $g(t)$ imply that $g(0) < 0$ , $g(1) > 0$ and $g (16) > 0$ . The first is satisfied by assumption since $g(0)=k <0$ , the second states that $(k+1)^2 > 0 \rightarrow k\neq -1$ and the third states that $0 > 16k^2 + 257 k + 16 =(16k+1)(k+16) \Rightarrow k < -16$ or $k > -\frac {1}{16}$ . Since we are restricted to $-5 < k < 0$ , the solution set is $-\frac {1}{16} < k < 0$ .

Case 3: $k=0$ . In this case, $f(x)=x^2$ doesn't satisfy the conditions.

Thus, the probability $p$ that $k$ satisfies the conditions is $p = \frac { \frac {1}{16}}{10} = 0.00625$ . Hence, $\lfloor 1000p \rfloor = 6$ .