Question

The number of real roots of the equation $e^{4x}+{2e}^{3x}-e^x-6=0$ is

• A. $1$
• B. $2$
• C. $4$
• D. $0$

Answer 1

Answer: (A)

Given: $e^{4x}+{2e}^{3x}-e^x-6=0$
Let $e^x=t$
$\Rightarrow t^4+2t^3-t-6=0$
$\Rightarrow t^4+2t^3+t^2-t^2-t-6=0$

$\Rightarrow (t^2{)}^2+2(t^2\left)\right(t)+(t{)}^2-t^2-t-6=0$
$\Rightarrow (t^2+t{)}^2-(t^2+t)-6=0$
Let $t^2+t=z$
$\Rightarrow z^2-z-6=0$
$\Rightarrow (z-3)(z+2)=0$
$\Rightarrow z=3\text{or}z=-2$
$\Rightarrow t^2+t-3=0\text{or}{t}^2+t+2=0\text{(rejected)}$
$\Rightarrow t=\frac{-1\pm \sqrt{13}}{2}$
Since, $e^x=t>0$
$\therefore e^x=\frac{-1+\sqrt{13}}{2}$
Hence, the given equation has only one real root.
Alternate Solution:
Let $e^x=t$
Now, assuming
$f\left(t\right)=t^4+2t^3-t-6,t>0$
$\Rightarrow f^{\prime }\left(t\right)=4t^3+6t^2-1$
$\Rightarrow f^{\prime \prime }\left(t\right)=12t^2+12t>0$$[\because t>0]$
So, $f^{\prime }\left(t\right)$ is always increasing.
Also, $f^{\prime }\left(0\right)=-1<0,f^{\prime }\left(1\right)=9>0$
So, $f^{\prime }\left(t\right)=0,$ for only one value of $t\in (0,1).$
Now, the nature of the graph is
$f\left(0\right)=-6<0$
$f\left(1\right)=-4<0$
$f\left(2\right)>0$

Hence, $f\left(t\right)=0$ has only one real solution.