Mathematics - Quadratic Equation Question with Solution | TestHub

Let$e^x=t\in \left(0,\infty \right)$
Given equation
$t^4+t^3-4t^2+t+1=0$
$t^2+t-4+\frac{1}{t}+\frac{1}{t^2}=0$
$\left(t^2+\frac{1}{t^2}\right)+\left(t+\frac{1}{t}\right)-4=0$
Let$t+\frac{1}{t}=\alpha$
$\left({\alpha }^2-2\right)+\alpha -4=0$
${\alpha }^2+\alpha -6=0$
${\alpha }^2+\alpha -6=0$
$\alpha =-3,2\Rightarrow \alpha =2\Rightarrow e^x+e^{-x}=2$
$x=0$is the only solution.