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

Question

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

Accepted Answer

${e}^{4x}+8{e}^{3x}+13{e}^{2x}-8{e}^{x}+1=0$

Let ${e}^{x}=t$

Now, ${t}^{4}+8{t}^{3}+13{t}^{2}-8t+1=0$

Dividing equation by ${t}^{2}$

$\begin{gathered} {t}^{2}+8t+13-\frac{8}{t}+\frac{1}{{t}^{2}}=0 \ {t}^{2}+\frac{1}{{t}^{2}}+8(t-\frac{1}{t})+13=0 \ {(t-\frac{1}{t})}^{2}+2+8(t-\frac{1}{t})+13=0 \end{gathered}$

Let $t-\frac{1}{t}=z$

${z}^{2}+8z+15=0$

$\begin{gathered} (z+3)(z+5)=0 \end{gathered}$

$z=-3z=-5$

So, $t-\frac{1}{t}=-3$ or $t-\frac{1}{t}=-5$

${t}^{2}+3t-1=0$ or ${t}^{2}+5t-1=0$

$t=\frac{-3\pm \sqrt{13}}{2}$ or $t=\frac{-5\pm \sqrt{29}}{2}$

as $t={e}^{x}$ so $t$ must be positive,

$t=\frac{\sqrt{13}-3}{2}\frac{\sqrt{29}-5}{2}$

So, $x=\ln (\frac{\sqrt{13}-3}{2})$ or $x=\ln (\frac{\sqrt{29}-5}{2})$

Hence two solutions and both are negative.

Solutions · 1