### Quadratic Equation Question 9

#### Question 9 - 31 January - Shift 2

The equation

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

(1) two solutions and both are negative

(2) no solution

(3) four solutions two of which are negative

(4) two solutions and only one of them is negative

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Roots of equations

$ 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 $t^2$,

$ \begin{aligned} & 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 \end{aligned} $

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

$ \begin{aligned} & z^2+8 z+15=0 \\ & (z+3)(z+5)=0 \\ & z=-3 \text { or } z=-5 \end{aligned} $

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

So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$

$ \text { or } \mathbf{x}=\ln \left(\frac{\sqrt{29}-5}{2}\right) $

Hence two solution and both are negative.