Show that $8x^4 −16x^3 +16x^2 −8x+k = 0$ has at least one non-real root for all real $k$. Find the sum of the non-real roots

Question

Show that $8x^4 −16x^3 +16x^2 −8x+k = 0$ has at least one non-real root for all real $k$. Find the sum of the non-real roots.

Since this polynomial looks so symmetric, I think factoring it might help. We have that $8(x^4-2x^3+2x^2-x) = 8x(x-1)(x^2-x+1)= -k$. Then I'm not sure how to work with the non-real root part, but I think that $(x^2-x+1)$ may have something to do with it.

Answer 1

No need to use derivatives, if you want to continue from your partial attempt.

$(x^2-x)(x^2-x+1) = (x^2-x+\frac{1}{2})^2-\frac{1}{4} = ((x-\frac{1}{2})^2+\frac{1}{4})^2-\frac{1}{4}$.

We can obviously see that it's decreasing for $x < \frac{1}{2}$ and increasing for $x > \frac{1}{2}$.

We certainly know more, since we want to find all $x$ such that:

$((x-\frac{1}{2})^2+\frac{1}{4})^2 = c$

where $c$ is some real.

$(x-\frac{1}{2})^2 = \pm\sqrt{c}-\frac{1}{4}$.

$x = \frac{1}{2} \pm \sqrt{\pm\sqrt{c}-\frac{1}{4}}$. [The signs are independent.]

All roots come in pairs with the outer "$\pm$". So the answer just depends on how many non-real roots. If you want to determine exactly when it has $2$ or $4$ non-real roots, just split into cases based on the $\sqrt{}$, giving the cases $c < 0$, or $0 \le c < \frac{1}{16}$, or $\frac{1}{16} \le c$. You may need to handle the boundaries separately if you don't want to count double roots as separate.

Answer 2

Hint To show that there is at least one nonreal root, we can observe the second derivative of the polynomial $p(x)$ on the l.h.s. is $$p''(x) = 32(3 x^2 - 3 x + 1);$$ this has discriminant $32^2 \cdot (-3) < 0$ and hence has the same sign for all $x$. What does this tell you about the number of real roots $p(x)$ can have?

For the sum, it's easier to work in the variable $x = u - \tfrac{1}{2}$. Then, $$p(x) = q(u) := 8 u^4 + 4 u^2 + \left(k - \tfrac{3}{2}\right) ,$$ which in particular is even. So, if $r$ is a root of $q(u)$, then so is $-r$, and hence $-\frac{1}{2} \pm r$ are roots of $p(x)$.

Answer 3

The first derivative of the root has $1$ real root and $2$ imaginary roots which are $0.5, 0.5+0.5i, 0.5-0.5i$. Hence, you can say that The equation can have only 2 real roots at the maximum. Since it is a fourth order equation, there must be at least $2$ imaginary roots.

Answer 4

The I-take-no-clever-shortcuts approach:

I consider the polynomial $p(x)=8x^4-16x^3+16x^2-8x$.

I try to rewrite $p(x)$ as something of the form $d(ax^2+bx+c)^2$ because if it is so, I know how to find the roots. The $d$ coefficient is there only because I like my leading coefficient of $x^4$ to be rational, and $8$ is not a square, but $8/2=4=2^2$ it is. So let me assume $d=2$, which is like to say I'm considering the coefficients of $p(x)/2$

Expanding $(ax^2+bx+c)^2$ I find $$a^2x^4 + 2 a b x^3 +(b^2 + 2 a c) x^2+2 b c x+c^2$$

it is not difficult to see if there is a choice (actually there are two with opposite signs) of $a,b,c$ that makes the above polynomial equal to the source one, minus the constant terms (that we will send to the right anyway).

Indeed we can immediately find that $a=\pm 2$. Let me try with $a=2$. From that I derive (from the coefficient of $x^3$) that $b=-2$ and then $c=1$.

All in all we have obtained $p(x)=2(2x^2-2x+1)^2 -2$ and thus that $p(x)=-k$ can be rewritten as $$ 2(2x^2-2x+1)^2 = 2-k $$

Now, it is not difficult to work out an expression for the roots of this equation. You simply consider the two equations (one for each sign) $$ 2x^2-2x+1 = \pm \sqrt{1-k/2} $$ and solve for $x$ using the classic formula. At the end you obtain four solutions, namely $$ x_{1,2,3,4}=\frac{1}{2}\left(1\pm\sqrt{-1\pm\sqrt{4-2k}}\right) $$

From here, it is easy to answer the questions of the problem.