Question

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

• A. four solutions two of which are negative
• B. two solutions and both are negative
• C. no solution
• D. two solutions and only one of them is negative

Answer 1

Answer: (B)

$e ^{4 x }+8 e ^{3 x }+13 e ^{2 x }-8 e ^{ x }+1=0$
Let $e ^{ x }= t$
Now, $t ^4+8 t ^3+13 t ^2-8 t +1=0$
Dividing equation by $t ^2$,
$ t ^2+8 t +13-\frac{8}{ t }+\frac{1}{ t ^2}=0 $
$ t ^2+\frac{1}{ t ^2}+8\left( t -\frac{1}{ t }\right)+13=0 $
$ \left( t -\frac{1}{ t }\right)^2+2+8\left( t -\frac{1}{ t }\right)+13=0$
Let $t -\frac{1}{ t }= z$
$ z^2+8 z+15=0 $
$ (z+3)(z+5)=0$
$z=-3$ or $z=-5$
So, $t-\frac{1}{t}=-3$ or $t-\frac{1}{t}=-5$
$t^2+3 t-1=0$ or $t^2+5 t-1=0$
$t=\frac{-3 \pm \sqrt{13}}{2}$ or $t =\frac{-5 \pm \sqrt{29}}{2}$
as $t=e^x$ so $t$ must be positive,
$t =\frac{\sqrt{13}-3}{2}$ or $\frac{\sqrt{29}-5}{2}$
So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}\right)$
Hence two solution and both are negative.