When does it cut/touch?

We are looking for the cases when $f(x)$ has a real double root, so that the graph touches the $x$ -axis. Since $x^4f(\frac{1}{x})=f(x)$ , if $c$ is a root, then so is $\frac{1}{c}$ . To get a double root, we want $c^2=1$ or $c=\pm 1$ . Now $f(\pm1)=1\pm2a$ so $a=\pm\frac{1}{2}$ and $a^2=\boxed{0.25}$ .

Method 1

Let $p$ be the double root and let $q \pm ri$ be the imaginary roots of $y = x^4 + ax^3 - x^2 + ax + 1$ . Then $(x - p)(x - p)(x - (q + ri))(x - (q - ri)) = 0$ or $x^4 - 2(p + q)x^3 + (4pq + p^2 + q^2 + r^2)x^2 - 2p(pq + q^2 + r^2)x + p^2(r^2 + q^2) = 0$ , which means $a = -2(p + q)$ , $4pq + p^2 + q^2 + r^2 = -1$ , $-2p(pq + q^2 + r^2) = a$ , and $p^2(r^2 + q^2) = 1$ .

Solving these equations for $p$ gives $p^6 + 4p^4 - 4p^2 - 1 = 0$ , which has real solutions $p = \pm 1$ . If $p = 1$ , then $q = -\frac{3}{4}$ , $r = \pm\frac{\sqrt{7}}{4}$ , and $a = \frac{1}{2}$ . If $p = -1$ , then $q = \frac{3}{4}$ , $r = \pm\frac{\sqrt{7}}{4}$ , and $a = -\frac{1}{2}$ . Either way, $a^2 = (\pm \frac{1}{2})^2 = \frac{1}{4} = \boxed{0.25}$ .

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Method 2

Let $p$ be the double root of $y = x^4 + ax^3 - x^2 + ax + 1$ . Then $(x - p)(x - p) = x^2 - 2px + p^2$ divides evenly into $x^4 + ax^3 - x^2 + ax + 1$ . Using long division we find that $\frac{x^4 + ax^3 - x^2 + ax + 1}{x^2 - 2px + p^2}$ has a remainder of $(4p^3 + 3ap^2 - 2p + a)x - (3p^4 + 2ap^3 - p^2 - 1)$ , so $4p^3 + 3ap^2 - 2p + a = 0$ and $3p^4 + 2ap^3 - p^2 - 1 = 0$ .

Solving these equations for $p$ gives $p^6 + 4p^4 - 4p^2 - 1 = 0$ , which has real solutions $p = \pm 1$ . If $p = 1$ , then $a = \frac{1}{2}$ . If $p = -1$ , then $a = -\frac{1}{2}$ . Either way, $a^2 = (\pm \frac{1}{2})^2 = \frac{1}{4} = \boxed{0.25}$ .

The solutions of $x^4+a x^3+a x-x^2+1=0$ are $\frac{\sqrt{a^2+12}}{4}-\frac{\sqrt{a^2-\frac{12 a}{\sqrt{a^2+12}}-\frac{a^3}{\sqrt{a^2+12}}-2}}{2 \sqrt{2}}-\frac{a}{4}, \\ \frac{\sqrt{a^2+12}}{4}+\frac{\sqrt{a^2-\frac{12 a}{\sqrt{a^2+12}}-\frac{a^3}{\sqrt{a^2+12}}-2}}{2 \sqrt{2}}-\frac{a}{4}, \\ -\frac{\sqrt{a^2+12}}{4}-\frac{1}{2} \sqrt{\frac{a^2}{2}-\frac{-a^3-12 a}{2 \sqrt{a^2+12}}-1}-\frac{a}{4}, \\ -\frac{\sqrt{a^2+12}}{4}+\frac{1}{2} \sqrt{\frac{a^2}{2}-\frac{-a^3-12 a}{2 \sqrt{a^2+12}}-1}-\frac{a}{4}$

Since $a^2+12>0\ \forall a$ , that discriminant is not the issue.

The solution of $a^2-\frac{12 a}{\sqrt{a^2+12}}-\frac{a^3}{\sqrt{a^2+12}}-2=0 \Longrightarrow a=-\frac12$ . and the solution of $\frac{a^2}{2}-\frac{-a^3-12 a}{2 \sqrt{a^2+12}}-1=0 \Longrightarrow a=\frac12$ . For the roots of the original equation to be real, the value of discriminants must be greater or equal to zero, therefore these are the values of $a$ in which we are interested. In either case, the square of the value is 0.25. Whether that is a minimum of the original equation is immaterial. Away from those values of $a$ , either the discriminant is less than 0 and the function is not real there or a is greater than zero and therefore is not at its minimum.

Divide the equation by x^2 and then, substitute (x+1/x) as t..............Now, in the quadratic, we need the condition that no roots lie between -2 and 2.........this gives the answer!!!!

Let $u=x+\frac{1}{x}$ . Then $u^2=x^2+\frac{1}{x^2}+2 \ge 2\sqrt{\left(x^2\right)\left(\frac{1}{x^2}\right)+2}=4$ (by AM-GM).

Now $f(x)=0$ implies that $\left(x^2+\frac{1}{x^2}\right)+a\left(x+\frac{1}{x}\right)-1 =0$ which means that $u^2+au-3=0$ . From this equation, $a=\frac{3}{u}-u$ and hence $a^2=\frac{9}{u^2}+u^2-6$ . As the function $\frac{9}{v}+v$ is increasing for all $v \ge1$ , $a^2=\frac{9}{u^2}+u^2-6 \ge \frac{9}{4}+4-6 =\boxed{0.25}$ .