6

I just wanted the solution, a hint or a start to the following question.

Determine the minimum of $a^2 + b^2$ if $a$ and $b$ are real numbers for which the equation $$x^4 + ax^3 + bx^2 + ax + 1 = 0$$ has at least one real solution.

Thanks in advance.

Start wearing purple
  • 53,234
  • 13
  • 164
  • 223
  • 2
    http://2000clicks.com/mathhelp/PuzzleReciprocalPolynomialAnswer.aspx – lab bhattacharjee Aug 23 '13 at 18:50
  • @labbhattacharjee thank you for the two links, very helpful – Shobhit Aug 23 '13 at 19:03
  • 1
    @Shivam, As the complement of at least one real root is all four roots are complex, I was trying to use http://www.jstor.org/stable/2972804 and http://en.wikipedia.org/wiki/Quartic_function#Nature_of_the_roots to find the condition(s) on $a,b$, but unfortunately the calculation appears to be too complex :) – lab bhattacharjee Aug 23 '13 at 19:07

3 Answers3

5

Hint: Your quartic is palindromic. Divide through by $x^2$. We get the equation $$x^2+ax+b+\frac{a}{x}+\frac{1}{x^2}=0.$$ Make the substitution $x+\frac{1}{x}=t$. Then $x^2+\frac{1}{x^2}=t^2-2$, and we arrive at the equation $$t^2+at+b-2=0.$$ Imagine solving the quadratic. We will get two not necessarily real values of $t$. The original equation has at least one real solution if and only if $t$ is real and has absolute value $\ge 2$. This is because, for example, for $x$ positive, the minimum value of $x+\frac{1}{x}$ is $2$.

Added, almost to the answers: There are two cases to consider, $a\ge 0$ and $a\lt 0$. Look at the case $a\ge 0$. The roots of the quadratic in $t$ are $$\frac{-a\pm \sqrt{a^2-4(b-2)}}{2}.$$ The root of largest absolute value is given by choosin the $-$ in $\pm$. We want this root to be $\le -2$. After some manipulation, that gives the inequality $$\sqrt{a^2-4(b-2)}\ge 4-a.$$ Squaring both sides and simplifying, we get $2a\ge b+ 2$. The minimum value of $a^2+b^2$ is attained when $2a=b+2$. That will give you the answer that has $a\ge 0$. The answer for negative $a$ is obtained by a similar calculation. The relevant root is then $t=\frac{-a+\sqrt{a^2-4(b-2)}}{2}$.

Remark: I feel somewhat guilty at bashing the problem with standard machinery. But routine calculation does work. There is undoubtedly a simple way. However, simple can take time.

André Nicolas
  • 507,029
  • 1
    The answer given is a=(+/-)\frac{4}5 and b=\frac{-2}5 and with your hints i cannot arrive at that. Please help –  Aug 23 '13 at 18:23
  • The answer given is $a$=$\pm\frac{4}5$ and $b=\frac{-2}5$ and with your hints i cannot arrive at that. Please help –  Aug 23 '13 at 18:42
  • 2
    The equation is also called Reciprocal equation (http://math.stackexchange.com/questions/403025/equation-with-high-exponents) – lab bhattacharjee Aug 23 '13 at 18:57
1

[The following solution relies on the ideas in the previous two answers.]

Let the roots of $x^4+ax^3+bx^2+ax+1=0$ be given by $c, \frac{1}{c}, d, \frac{1}{d}$ where $c\in\mathbb{R}$, and

let $t=c+\frac{1}{c}$ and $l=d+\frac{1}{d}$.

Then $|t|\ge2$ since $c+\frac{1}{c}\ge2$ if $c>0$ and $c+\frac{1}{c}\le2$ if $c<0$.

Since $x^4+ax^3+bx^2+ax+1=(x-c)(x-\frac{1}{c})(x-d)(x-\frac{1}{d})=(x^2-tx+1)(x^2-lx+1)$,

we have that $a=-(t+l)$ and $b=2+lt$; so

$a^2+b^2=(t+l)^2+(2+lt)^2=t^2+2lt+l^2+4+4lt+l^2t^2=(l^2+1)t^2+6lt+l^2+4$.

We consider two cases:

$\mathbf{A)}$ If $d\in\mathbb{R}$, then $|l|\ge2$ (for the same reason that $|t|\ge2$); so

$\;\;\;\;a^2+b^2=(l^2+1)t^2+6lt+l^2+4\ge20+6lt+8\ge4$ $\;$(since $|lt|\le4\implies lt\ge-4$).

$\mathbf{B)}$ If $d\not\in\mathbb{R}$, then $\frac{1}{d}=\bar{d}\implies|d|=1$, so $d=e^{i\theta}=\cos\theta+i\sin\theta$ for some $\theta\in\mathbb{R}$. Then

$\;\;\;\;l=d+\frac{1}{d}=d+\bar{d}=2\cos\theta$, so $|l|\le2$.

With $l$ fixed, let $f(t)=a^2+b^2=(l^2+1)t^2+6lt+l^2+4$.

Since the vertex of the graph of $f$ is at $t=\frac{-3l}{l^2+1}$,

and $-2<\frac{-3l}{l^2+1}<2\;\;\;\;$ (since $2l^2+3l+2>0$ and $2l^2-3l+2>0$),

the minimum value of f for $|t|\ge2$ will occur when $t=2$ or $t=-2$:

1) If $t=2$, $a^2+b^2=4(l^2+1)+12l+l^2+4=5l^2+12l+8$, and

$\;\;\;g(l)=5l^2+12l+8$ has its minimum when $l=-\frac{6}{5}$, with $g(-\frac{6}{5})=\frac{4}{5}$.

$\;\;\;\;$(Notice that $|-\frac{6}{5}|\le2$).

2) If $t=-2$, $a^2+b^2=4(l^2+1)-12l+l^2+4=5l^2-12l+8$, and

$\;\;\;g(l)=5l^2-12l+8$ has its minimum when $l=\frac{6}{5}$, with $g(\frac{6}{5})=\frac{4}{5}$.

$\;\;\;\;$(Notice that $|\frac{6}{5}|\le2$).

From parts A and B, we can conclude that the minimum value of $a^2+b^2$ is $\;\;4/5$.

user84413
  • 27,211
0

As has already been observed, if $c$ is a root of your equation, so is $c^{-1}$. So suppose your equation has roots $c,d,c^{-1},d^{-1}$.

We have $$a=-(c+c^{-1} + d + d^{-1})$$ and $$b=cc^{-1}+ cd+cd^{-1} +c^{-1}d +c^{-1}d^{-1} + dd^{-1}$$ $$=2+(c+c^{-1})(d+d^{-1})$$

Now if $c\in \mathbb{R}$, we must also have $d+d^{-1}\in \mathbb{R}$. This means $d=i$ or $d\in \mathbb{R}$.

Thus if we write $z=c+c^{-1}$ and $w=d + d^{-1}$, we see that your problem is a question of minimising $$f=(w+z)^2 + (2+wz)^2$$ subject to the constraint $|w|\geq 2$ and $|z|\geq 2$ or $z=0$.

If $z=0$ the $f=w^2+4\geq8$. Consider instead $|z|\geq 2$. In this case $|zw|\geq 4$ so $f\geq (2-zw)^2\geq 4$. Thus we see that the minimal value of $f$ is 4. This is obtained when $z=2,w=-2$. And the equation is $$x^4-2x^2+1$$