Number of real solution of $e^{4x}+8e^{3x}+13e^{2x}-8e^x+1=0$
What I try
Using Aithmetu geometric Inequality
$\displaystyle e^{4x}+8e^{3x}+13e^{2x}+1\geq 4(104e^{9x})^{\frac{1}{4}}>8e^x$
So $e^{4x}+8e^{3x}+13e^{2x}-8e^x+1>0$ for all real $x$
But answer given as $2$ real solution
Plz have a look on my problem and what's wrong in this solution
Hint: Set $z=e^x$. This gives
$$z^4+8z^3+13z^2-8z+1 \, = \, \left(z^2 + 3 z - 1\right) \left(z^2 + 5 z - 1\right) \, =\, 0$$
Each quadratic corresponds to two zeroes, for a total of four $z$ values. Finally, you can take their natural log, to find the four corresponding $x$ values. All four $z$ values are real, and the two positive ones correspond to real $x$ values.
The error in your answer here :
$$4(104e^{9x})^{\dfrac{1}{4}}> 8e^x$$
put $ x=-4$ $$4(104e^{9x})^{\dfrac{1}{4}}= 4(104e^{-36})^{\dfrac{1}{4}}=0.001576... $$
$$8e^x=8e^{-4}=0.1465..$$
Let $u=e^x$ so we have
$u^4+8u^3+13u^2-8u+1=0$
$(u^2+3u-1)(u^2+5u-1)=0$
Applying the quadratic formula to each factor we get
$u=\frac{-3\pm\sqrt{13}}{2}$ and $u=\frac{-5\pm\sqrt{29}}{2}$
Since $u=e^x$,we eliminate the negative values and are left with two possibilities.
$e^x=\frac{-3+\sqrt{13}}{2} $ and $e^x=\frac{-5+\sqrt{29}}{2}$.
You can solve for $x$ if you like but this is sufficient to show you will have two real solutions.
