Shourya's non-real roots

First, we consider $f(y)=128y^4-128y^3+64y^2-16y$ , and $\frac{df}{dy}=512y^3-384y^2+128y-16$ .

$\frac{df}{dy}=0$ only when $y=\frac{1}{4}$ , and at this point, $f(y)=\frac{-3}{2}$ . We have two cases:

Case 1: $r>\frac{3}{2}$ . In this case, $f(y)+r >0\forall y$ , so all roots of $f(y)+r=0$ are complex and according to vieta's theorem, $X_r=-\frac{-128}{128}=1$ .

Case 2: $r\leq \frac{3}{2}$ . In this case, since $\frac{df}{dy}=0$ has only 1 real root, the equation $f(y)+r=0$ has two real roots (multiple root when $r=\frac{3}{2}$ is equivalent).

When r=0, we can easily find two real roots of $f(y)+r=0$ : y=0 and $y=\frac{1}{2}$ .

Because of the symmetry of f(y), when r changes (but still satisfies $r\leq \frac{3}{2}$ ), the sum of these real roots do not change. Therefore, in this case, $X_r=1-(0+\frac{1}{2})=\frac{1}{2}$ .

Therefore, the sum of all possible distinct values of $X_r$ is $1+\frac{1}{2}=\frac{3}{2}$ .

The answer is: 3+2=5.

Let $f_r(y) = 128y^4-128y^3+64y^2-16y+r$ . We shall write $f_r(y)$ in the form of $p(y^2+qy+r)^2+sy+t$ . It turns out that $f_r(y) = 128(y^2-\frac{1}{2}y+\frac{1}{8})^2+r-2$ . Now we can complete the square for the quadratic expression inside the brackets to get $f_r(y) = 128[(y-\frac{1}{4})^2+\frac{1}{16}]^2+r-2$ . Now we try to solve for the roots of $f_r(y) = 0$ . The roots of $f_r(y) = 0$ satisfy $[(y-\frac{1}{4})^2+\frac{1}{16}]^2 = \frac{2-r}{128}$ . Clearly if $r > 2$ , there are no real roots, so there are $4$ non-real roots (counting multiplicity). Since non-real roots of polynomials with real coefficients come in conjugates, we either have all four roots distinct, or we have a pair of conjugate roots (both with multiplicity $2$ ). When all four roots are distinct, Vieta's formulae give the sum of the roots as $-\frac{-128}{128} = 1$ . Otherwise, we have counted each root twice, so the sum is half of the previous case, which is $\frac{1}{2}$ . Now consider $r \leq 2$ . Then we must have $(y-\frac{1}{4})^2 = - \sqrt{\frac{2-r}{128}} - \frac{1}{16}$ or $(y-\frac{1}{4})^2 = \sqrt{\frac{2-r}{128}} - \frac{1}{16}$ . The first equation has clearly no real solutions and gives rise to $2$ non-real conjugate roots for $y$ . The second equation gives no real solution when $\sqrt{\frac{2-r}{128}} < \frac{1}{16}$ , so this means there will be a total of four non-real roots (counting multiplicity) and we have the same scenario as when $r > 2$ . But when $\sqrt{\frac{2-r}{128}} \geq \frac{1}{16}$ , then the roots will be real, so the only non-real roots would come from the first equation $(y-\frac{1}{4})^2 = - \sqrt{\frac{2-r}{128}} - \frac{1}{16}$ , and since the coefficients of $y^2$ and $y$ are $1$ and $-\frac{1}{2}$ respectively, we know that the sum of the two non-real conjugate roots (which have to be distinct) is $\frac{1}{2}$ by Vieta's formulae.

It is easy to find $r$ such that $X_r = \frac{1}{2}$ and $X_r = 1$ respectively. For instance, if we set $r = 0$ , we have $\sqrt{\frac{2-r}{128}} \geq \frac{1}{16}$ , so by the above reasoning, we must have $X_0 = \frac{1}{2}$ . On the other hand, if we set $r = 146$ , then $\sqrt{\frac{2-r}{128}} = \sqrt{-\frac{9}{8}} = \pm \frac{3\sqrt{2}}{4}i$ , so we have either $(y-\frac{1}{4})^2 = \pm \frac{3\sqrt{2}}{4}i - \frac{1}{16}$ . Observe that since $(\frac{\sqrt{2}}{2} \pm \frac{3}{4}i)^2 = \pm \frac{3\sqrt{2}}{4}i - \frac{1}{16}$ , we have $y - \frac{1}{4} = \frac{\sqrt{2}}{2} + \frac{3}{4}i$ or $y - \frac{1}{4} = - (\frac{\sqrt{2}}{2} + \frac{3}{4}i)$ or $y - \frac{1}{4} = \frac{\sqrt{2}}{2} - \frac{3}{4}i$ or $y - \frac{1}{4} = -(\frac{\sqrt{2}}{2} - \frac{3}{4}i)$ . Each of these results in a distinct root, so we know that it is possible to have $4$ non-real roots that sum to $1$ , i.e. $X_{146} = 1$ . Therefore, $S = 1 + \frac{1}{2} = \frac{3}{2}$ , and our answer is $\boxed{5}$ .

[Of course, there are easier ways to show that $X_r = 1$ is attainable; the above is just showing one explicit example.]

This is my first time of writing a solution, so forgive me for any mistakes.<P> Denote y=128x^4−128x^3+64x^2−16x+r and observe the function,it is a parabola which is symmetrical to x=1/4 and has a minimum value of y if and only if x=1/4.(It's not hard to calculate this by doing some differentiation but I'll skip here.) First, substitute x=1/4 into y and let y=0.(I'm doing this to find the value of r when the minimum point of y is (1/4,0).) Obviously, r=3/2. For all r which is greater than 3/2, y has no real roots as the curve is above x-axis. According to Vieta's theorem, the sum of all non-real roots of the equation is equal to -(-128)/128=1. Now, we consider the other circumstances when r is smaller than or equal to 3/2. Under this circumstance, y has one real root(when r=3/2) or two real roots which are symmetrical to x=1/4 (when r<3/2). So, we let both real roots be 1/4+d and 1/4-d (d > 0). Then,we divide 128x^4−128x^3+64x^2−16x+r by (x-(1/4+d))(x-(1/4-d)). We will get 128x^2-64x+C where C is a function of r. Clearly there are no real roots for 128x^2-64x+C=0.(Since the equation of y has only two real roots and now we do the division to eliminate the real roots from y.) Therefore, the sum of all non-real roots of the equation 128x^4−128x^3+64x^2−16x+r=0 for all r is smaller than or equal to 3/2 is 64/128=1/2(according to Vieta's theorem again). 1+1/2=3/2. Therefore 3+2=5.