Question

The number of real solutions of the equation e^4x + 4e^3x - 58e^2x + 4ex + 1 = 0 is _____.

Answer 1

Correct Answer: 2

The given equation e^4x + 4e^3x - 58e^2x + 4ex + 1 = 0
Dividing by e^2x
e^2x + 4e^x – 58 + 4e^–x + e^–2x = 0
⇒ (e^x + e^–x)² + 4(e^x + e^–x) – 60 = 0
Let e^x + e^–x = t ∈ [2, ∞)
⇒ t² + 4t – 60 = 0
⇒ t = 6 is only possible solution
e^x + e^–x = 6
⇒ e^2x – 6e^x + 1 = 0
Let e^x = p,
p² – 6p + 1 = 0
\(⇒ p = \frac {3+\sqrt 5}{2}\) or,  \(\frac {3-\sqrt 5}{2}\)
Therefore,
\(x = ln (\frac {3+\sqrt 5}{2})\)

or, \(x=ln (\frac {3-\sqrt 5}{2} )\)

So, The number of real solutions of the equation is 2.