Question

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

Hint:

For equations symmetric in $e^x$ and $e^{-x}$, always try the substitution $y=e^x$ and check domain restrictions before accepting roots.

Answer 1

Correct Answer: 2

Given equation: \[ e^{4x}-e^{3x}-4e^{2x}-e^x+1=0 \] Let \[ y=e^x \quad (y > 0) \] Substituting, \[ y^4-y^3-4y^2-y+1=0 \] This is a reciprocal equation of the form: \[ ay^4+by^3+cy^2+by+a=0 \] Since \( y \neq 0 \), divide throughout by \( y^2 \): \[ y^2-y-4-\frac{1}{y}+\frac{1}{y^2}=0 \] Rearranging, \[ \left(y^2+\frac{1}{y^2}\right)-\left(y+\frac{1}{y}\right)-4=0 \] Let \[ z=y+\frac{1}{y} \Rightarrow y^2+\frac{1}{y^2}=z^2-2 \] Substitute: \[ (z^2-2)-z-4=0 \Rightarrow z^2-z-6=0 \] \[ (z-3)(z+2)=0 \Rightarrow z=3 \text{ or } z=-2 \] Since \( y > 0 \), by AM–GM: \[ y+\frac{1}{y} \ge 2 \] Hence \( z=-2 \) is rejected. So, \[ y+\frac{1}{y}=3 \] Multiplying by \( y \): \[ y^2-3y+1=0 \] \[ y=\frac{3\pm\sqrt5}{2} \] Both roots are positive, hence both are valid. Since \( y=e^x \), each value of \( y \) gives one real value of \( x \). \[ \boxed{\text{Number of real roots }=2} \]