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

Put $$y = e^{x}$$. Because $$e^{x}\gt 0$$ for every real $$x$$, the given equation becomes a quartic in the single positive variable $$y$$:

$$y^{4}+8y^{3}+13y^{2}-8y+1 = 0 \qquad (1)$$

Try to split the quartic into two quadratics of the form$$(y^{2}+ay-1)(y^{2}+by-1).$$Multiplying we obtain

$$y^{4}+(a+b)y^{3}+(ab-2)y^{2}-(a+b)y+1.$$(1)

Comparing the coefficients of (1) with those in equation (1) gives the system

$$a+b = 8,\qquad ab-2 = 13,\qquad -(a+b) = -8.$$

The first and third equalities are identical. From $$a+b=8$$ and $$ab-2=13$$ we have $$ab = 15$$. Hence $$a,b$$ satisfy $$t^{2}-8t+15=0$$, whose roots are $$t=5,3$$. Thus

$$y^{4}+8y^{3}+13y^{2}-8y+1=(y^{2}+5y-1)(y^{2}+3y-1)=0.$$

Solve each quadratic separately.

$$y^{2}+5y-1=0\;\Rightarrow\;y=\frac{-5\pm\sqrt{25+4}}{2} =\frac{-5\pm\sqrt{29}}{2}.$$

Since $$\sqrt{29}\approx5.385$$,  $$y_{1}=\frac{-5+\sqrt{29}}{2}\approx0.1925\;(\gt0),\qquad y_{2}=\frac{-5-\sqrt{29}}{2}\approx-5.1925\;(\lt0).$$

$$y^{2}+3y-1=0\;\Rightarrow\;y=\frac{-3\pm\sqrt{9+4}}{2} =\frac{-3\pm\sqrt{13}}{2}.$$

Because $$\sqrt{13}\approx3.605$$,  $$y_{3}=\frac{-3+\sqrt{13}}{2}\approx0.3025\;(\gt0),\qquad y_{4}=\frac{-3-\sqrt{13}}{2}\approx-3.3025\;(\lt0).$$

Only the positive roots satisfy $$y=e^{x}\gt0$$, so from $$y_{1}$$ and $$y_{3}$$ we get

$$x_{1}=\ln y_{1}=\ln\!\left(\frac{-5+\sqrt{29}}{2}\right),\quad x_{2}=\ln y_{3}=\ln\!\left(\frac{-3+\sqrt{13}}{2}\right).$$

Both numbers $$y_{1},y_{3}$$ are between 0 and 1, hence $$\ln y_{1}\lt0$$ and $$\ln y_{3}\lt0$$. Therefore

• exactly two real values of $$x$$ satisfy the original equation,
• and both of those values are negative.

Option B which is: two solutions and both are negative.