Question

The number of real roots of the equation 

e^4x - e^3x - 4e^2x - e^x + 1 = 0 is equal to____ .

Answer 1

t⁴ – t³ – 4t² – t + 1 = 0, e^x = t > 0 

\(\Rightarrow\) t² – t – 4 - \(\frac{1}{t} + \frac{1}{t^2} = 0\)

\(\Rightarrow\) α² - α - 6 = 0, α = t + \(\frac{1}{t} \ge 2\)

\(\Rightarrow\) α =3, -2 (reject)

\(\Rightarrow\) t + \(\frac{1}{t}\) = 3 

\(\Rightarrow\) The number of real roots = 2