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Problem Of The Week #416 May 11th, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,516
Here is this week's POTW:

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Show that for every real number $a$ the equation $8x^4-16x^3+16x^2-8x+a=0$ has at least one non-real root and find the sum of all the non-real roots of the equation.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,516
Congratulations to topsquark for his correct solution(Cool), which you can find below:

Solution from topsquark :
\(\displaystyle 8x^4 - 16x^3 + 16x^2 - 8x + a\)

To make the numbers a tad easier I'm going to define b = (1/8) a and divide the above expression by 8. This does not affect the zeros of the polynomial in any way.

Let
\(\displaystyle f(x) = x^4 - 2 x^3 + 2 x^2 - x\)

Then
\(\displaystyle f'(x) = 4x^3 - 6 x^2 + 4x - 1\)
and
\(\displaystyle f''(x) = 12 x^2 - 12 x + 4\)

Now, the critical point of f(x) is x = 1/2. Putting this into the second derivative we get that x = 1/2 is a relative minimum. But there is only one critical point so we know that x = 1/2 is an absolute minimum and it's the only critical point for f(x) so f(x) looks a bit like a parabola. b merely moves the graph of the original polynomial vertically so it has two real solutions or no real solutions. Either way it has a complex zero.

For the sum of the complex zeros I found that we can actually solve the quartic.
\(\displaystyle x^4 - 2 x^3 + 2x^2 - x + b = 0\)

Let's find the depressed quartic. Let y = x - 1/2. Then
\(\displaystyle y^4 + (1/2) y^2 + (b - 3/16) = 0\)

Surprisingly this is a biquardratic. So solving this in the usual way I get that
\(\displaystyle y = \sqrt{-1/4 \pm \sqrt{1/4 - b}}\) (where the \(\displaystyle \pm\)'s are independent of each other.)

Finally, after some simplifying and factoring we get:
\(\displaystyle x = (1/2) (1 \pm \sqrt{1 \pm 2 \sqrt{1 - 4b}}\)

or
\(\displaystyle x = \begin{cases} (1/2) (1 \pm \sqrt{1 + 2 \sqrt{1 - 4b}}) \\ (1/2) (1 \pm \sqrt{1 - 2 \sqrt{1 - 4b}}) \end{cases}\)

Notice that the second case is complex for both \(\displaystyle \pm\)'s. Thus this represents the two complex solutions. If there are only two complex solutions, the sum of these will be
\(\displaystyle (1/2) (1 \pm \sqrt{1 + 2 \sqrt{1 - 4b}}) + (1/2) (1 - \sqrt{1 \pm 2 \sqrt{1 - 4b}}) = (1/2) \cdot 1 + (1/2) \cdot 1 = 1\)

If all four zeros are complex then the sum of these will be
\(\displaystyle (1/2) (1 + \sqrt{1 + 2 \sqrt{1 - 4b}}) + (1/2) (1 - \sqrt{1 + 2 \sqrt{1 - 4b}}) + (1/2) (1 + \sqrt{1 - 2 \sqrt{1 - 4b}}) + (1/2) (1 - \sqrt{1 - 2 \sqrt{1 - 4b}}) = (1/2) \cdot 1 + (1/2) \cdot 1 + (1/2) \cdot 1 + (1/2) \cdot 1 = 2\).

We can also show for what values of b (and thus a) will give two complex or four complex zeros. Look at the real solutions for x. Note that for there to be only one real zero these two must be equal. This means that
\(\displaystyle x = (1/2) (1 + \sqrt{1 + 2 \sqrt{1 - 4b}}) = (1/2) (1 - \sqrt{1 + 2 \sqrt{1 - 4b}})\)

The only way this can happen is for the square roots to be 0:
\(\displaystyle \sqrt{1 + 2 \sqrt{1 - 4b}}) = 0 \implies b = 3/16 \implies a = 3/2\). So for \(\displaystyle a > 3/2\) we get four complex zeros and for \(\displaystyle a \le 3/2\) we get two real and two complex zeros.

Alternative solution from other:
Substituting $x=y+\dfrac{1}{2}$ in the equation we obtain an equation in $y$:

$8y^4+4y^2+a-\dfrac{3}{2}=0$

Using the transformation $z=y^2$, we get a quadratic equation in $z$:

$8z^2+4z+a-\dfrac{3}{2}=0$

The discriminant of this equation is $32(2-a)$, which is non-negative if and only if $a\le 2$. For $a\le 2$, we obtain the roots

$z_1=\dfrac{-1+\sqrt{2(2-a)}}{4},\,z_2=\dfrac{-1-\sqrt{2(2-a)}}{4}$

For getting real $y$ we need $z\ge 0$. Obviously $z_2<0$ and hence it gives only non-real values of $y$. But $z_1\ge 0$ if and only if $a\le \dfrac{3}{2}$. In this case we obtain two real values for $y$ and hence two real roots for the original equation.

Thus we conclude that there are two real roots and two non-real roots for $a\le \dfrac{3}{2}$ and four non-real roots for $a>\dfrac{3}{2}$.

Obviously the sum of all the roots of the equation is 2. For $a\le \dfrac{3}{2}$. two real roots of $8y^4+4y^2+a-\dfrac{3}{2}=0$ are given by $y_1=\sqrt{z_1}$ and $y_2=-\sqrt{z_1}$. Hence the sum of real roots of the original equation is given by $y_1+\dfrac{1}{2}+y_2+\dfrac{1}{2}$ which reduces to 1. It follows the sum of the non-real roots of (1) for $a\le \dfrac{3}{2}$ is also 1. Thus,

$
\text{The sum of non-real roots} =
\begin{cases}
1 & \text{for $a \le \dfrac{3}{2}$} \\
2 & \text{for $a<\dfrac{3}{2}$} \\
\end{cases}$
 
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