Question

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

• A. two solutions and both are negative
• B. two solutions and only one of them is negative
• C. four solutions two of which are negative
• D. no solution

Hint:

Put \(t = e^x > 0\) and factorise the quartic as \((t^2+3t-1)(t^2+5t-1)\); count the positive roots.

Answer 1

The Correct Option is A

Let \(t = e^{x}\), so \(t > 0\). The equation becomes \(t^4 + 8t^3 + 13t^2 - 8t + 1 = 0\).
Factorise: \((t^2 + 3t - 1)(t^2 + 5t - 1) = t^4 + 8t^3 + (15 - 2)t^2 - 8t + 1 = t^4 + 8t^3 + 13t^2 - 8t + 1\). Correct.
\(t^2 + 3t - 1 = 0 \Rightarrow t = \frac{-3 \pm \sqrt{13}}{2}\); only \(t = \frac{-3 + \sqrt{13}}{2} \approx 0.303 > 0\) is admissible.
\(t^2 + 5t - 1 = 0 \Rightarrow t = \frac{-5 \pm \sqrt{29}}{2}\); only \(t = \frac{-5 + \sqrt{29}}{2} \approx 0.193 > 0\) is admissible.
Both positive values of \(t\) lie in \((0, 1)\), so \(x = \ln t < 0\) for both.
Hence there are exactly two real solutions and both are negative.
Answer: (1) two solutions and both are negative