Multiple choice

The number of real roots of the equation, $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$ is?

  1. $3$
  2. $4$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let u = e^x. The equation becomes u^4 + u^3 - 4u^2 + u + 1 = 0. Dividing by u^2 (since u > 0): u^2 + u - 4 + 1/u + 1/u^2 = 0. (u^2 + 1/u^2) + (u + 1/u) - 4 = 0. Let t = u + 1/u. Then u^2 + 1/u^2 = t^2 - 2. (t^2 - 2) + t - 4 = 0 => t^2 + t - 6 = 0 => (t+3)(t-2) = 0. Since u > 0, t = u + 1/u >= 2. Thus t = 2. u + 1/u = 2 => u = 1. e^x = 1 => x = 0. Only one real root.

AI explanation

Divide the equation by e^(2x) to get y^2 + y - 4 + 1/y + 1/y^2 = 0, where y = e^x. Substituting t = y + 1/y transforms the equation into t^2 - 2 + t - 4 = 0, yielding t^2 + t - 6 = 0. Factoring gives (t - 2)(t + 3) = 0, so the valid root is t = 2 since t must be at least 2 or at most -2. Solving y + 1/y = 2 results in y = 1, which means e^x = 1 and produces exactly one real root at x = 0.

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