Question

The number of real roots of the equation, $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$ is

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

The correct option is C

$1$

Explanantion for the correct option:

Find the roots of the given equation.

An equation $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$ is given.

Solve the given equation as follows:

Divide both sides by $e^x$.

$$\begin{gathered} e^{2x}+e^x-4+\frac{1}{e^x}+\frac{1}{e^{2x}}=0 \\ \Rightarrow \left(e^{2x}+\frac{1}{e^{2x}}\right)+\left(e^x+\frac{1}{e^x}\right)-4=0 \\ \Rightarrow {\left(e^x+\frac{1}{e^x}\right)}^2-2+\left(e^x+\frac{1}{e^x}\right)-4=0 \end{gathered}$$

Assume that, $e^x+\frac{1}{e^x}=t$.

$$\begin{gathered} t^2+t-6=0 \\ \Rightarrow t=\frac{-1\pm \sqrt{1-4(-6)}}{2} \\ \Rightarrow t=\frac{-1\pm \sqrt{1+24}}{2} \\ \Rightarrow t=\frac{-1\pm \sqrt{25}}{2} \\ \Rightarrow t=\frac{-1\pm 5}{2} \\ \Rightarrow t=-3,2 \end{gathered}$$

Since, $e^x+\frac{1}{e^x}=t$ So, $t$ cannot be negative.

Therefore, $t=2$.

Find the value of $\mathrm{x}$.

$e^x+\frac{1}{e^x}=2$

Therefore, $x=0$ is the only solution.

Thus, The number of real roots of the equation, $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$ is $1$.

Hence, option $C$ is the correct answer.