$|a^{2x}+a^{x+2}-1|\ge 1$ equation for positive a

Question

If $|a^{2x}+a^{x+2}-1|\ge 1$ for all values of a(a>0), $\ne 1$. Find the domain of x.

I tried to substitute $a^x=t$ and used the following $| t^{2}+a^2t-1|\ge 1$ but it is getting complicated.

Answer 1

The answer is $-2<x<0$. First note that $|a^{2x}+a^{x+2}-1| \geq 1$ iff $a^{2x}+a^{x+2}-1 \geq 1$ or $a^{2x}+a^{x+2}-1 \leq -1$. But the second possibility is ruled out because $a^{2x}+a^{x+2}$ is always positive. Hence we have to find $x$ such that $a^{2x}+a^{x+2} \geq 2$. If $x\geq 0$ then we get a contradiction to the inequality by letting $a$ decrease to $0$. Hence $x<0$. If $x\leq -2$ we get a contradiction by letting $a \to \infty$. Hence $-2<x<0$. To prove that the inequality holds for all $x$ in this range for all $a>0$ you can check that $a^{2x}+a^{x+2}$ is a decreasing function of $a$ in $(0,1)$ and an increasing function in $(1,\infty )$. [For this check that both terms are decreasing when $a<1$ and both terms are increasing when $a>1$]. Hence its minimum value with respect to $a$ is attained at $a=1$. Since $a^{2x}+a^{x+2} \geq 2$ is satisfied when $a=1$ we see that it holds for all $a>0$.

Answer 2

Assuming that $a, t > 0$,

If $t^2 + a^2 t - 1 < 0$, then

\begin{align} |t^2 + a^2 t - 1| &\ge 1 \\ 1 - a^2 t - t^2 &\ge 1 \\ t^2 + a^2 t &< 0 \end{align}

Which has no positive real solution.

If $t^2 + a^2 t - 1 \ge 0$, then

\begin{align} |t^2 + a^2 t - 1| &\ge 1 \\ t^2 + a^2 t - 1 &\ge 1 \\ (t + \frac 12 a^2)^2 &\ge 2 + \frac 14 a^4 \\ t + \frac 12 a^2 &\ge \frac 12\sqrt{8 + a^4} \\ t &\ge \frac 12\left(\sqrt{8 + a^4} - a^2 \right) \\ t &\ge \dfrac{4}{\sqrt{8 + a^4} + a^2} \\ a^x &\ge \dfrac{4}{\sqrt{8 + a^4} + a^2} \\ x \ln a &\ge \ln 4 -\ln(\sqrt{8 + a^4} + a^2) \\ x &\ge \dfrac{\ln 4 - \ln(\sqrt{8 + a^4} + a^2)}{\ln a} \\ &\text{OR} \\ a^x &\ge \frac 12\left(\sqrt{8 + a^4} - a^2 \right) \\ x \ln a &\ge \ln{\left(\sqrt{8 + a^4} - a^2 \right)} - \ln 2 \\ x &\ge \dfrac{\ln{\left(\sqrt{8 + a^4} - a^2 \right)}}{\ln a} \end{align}