The number of real solutions of the equation e⁴ˣ + 4e³ˣ - 58e²ˣ + 4eˣ + 1 = 0 is ___________.

Solution

Maths sir · 4 steps

The coefficients read the same from both ends, so divide by e²ˣ and set y = eˣ + e⁻ˣ.

1. Substitute and divide. Put t = eˣ, where t > 0. The equation becomes t⁴ + 4t³ - 58t² + 4t + 1 = 0. Its coefficients 1, 4, −58, 4, 1 are symmetric, so divide by t² to get (t² + 1/t²) + 4(t + 1/t) - 58 = 0.
2. Quadratic in y. Let y = t + 1/t. Then t² + 1/t² = y² - 2. Since t > 0, AM–GM gives y ≥ 2.

   y² + 4y - 60 = 0 ⇒ (y + 10)(y - 6) = 0

3. Keep valid root. y = -10 is impossible because y ≥ 2, so y = 6. Then t² - 6t + 1 = 0, which gives t = 3 ± 2√2. Both values are positive.
4. So. Each positive t gives exactly one real x = ln t, namely x = ln(3 + 2√2) and x = ln(3 - 2√2). The number of real solutions is 2.

The common slip: 4. Counting all four roots of the quartic in t, including the two negative ones from y = -10, gives 4. But eˣ can never be negative.

The idea to keep. When the coefficients are symmetric, divide by the middle power and use y = t + 1/t. After that, check every root against the condition y ≥ 2 that comes from t = eˣ > 0.

Answer from the NTA final answer key. Solution by PrepWiser.