The equation e^4x+8e^3x+13e^2x-8e^x+1=0, x R, has:

Question

The equation $e^{4x}+8e^{3x}+13e^{2x}-8e^x+1=0,\ x\in\mathbb R$, has:

Accepted Answer

Let $t=e^x>0$. Dividing by $t^2$:
$$t^2+\frac1{t^2}+8\left(t-\frac1t\right)+13=0.$$

Let $z=t-\dfrac1t$, so $t^2+\dfrac1{t^2}=z^2+2$:
$$z^2+8z+15=0 \implies (z+3)(z+5)=0 \implies z=-3 \text{ or } z=-5.$$

Case $z=-3$: $t^2+3t-1=0 \implies t=\dfrac{-3+\sqrt{13}}2$ (positive root; the other root is negative and rejected).

Case $z=-5$: $t^2+5t-1=0 \implies t=\dfrac{-5+\sqrt{29}}2$ (positive root).

Both values of $t$ lie strictly between 0 and 1 (since $\sqrt{13}\approx3.6$ gives $t\approx0.30$, and $\sqrt{29}\approx5.4$ gives $t\approx0.19$), so $x=\ln t<0$ in both cases.

Hence there are two solutions, both negative.

Solutions · 1