Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

1

Answer 2

Let the given equation be $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$. Let $y = e^x$. Since $x$ is a real number, $e^x > 0$, so $y > 0$. Substituting $y$ into the equation, we get: $y^4 + y^3 - 4y^2 + y + 1 = 0$

This is a quartic equation in $y$. We are looking for the number of positive real roots of this equation, because each positive real root $y$ corresponds to a unique real root $x = \ln y$ for the original equation.

The equation is a reciprocal equation since the coefficients (1, 1, -4, 1, 1) are symmetric. Since $y=0$ is not a root, we can divide the equation by $y^2$: $y^2 + y - 4 + \frac{1}{y} + \frac{1}{y^2} = 0$ Grouping terms: $(y^2 + \frac{1}{y^2}) + (y + \frac{1}{y}) - 4 = 0$

Let $z = y + \frac{1}{y}$. Squaring $z$, we get $z^2 = (y + \frac{1}{y})^2 = y^2 + 2 + \frac{1}{y^2}$. So, $y^2 + \frac{1}{y^2} = z^2 - 2$.

Substitute this into the equation: $(z^2 - 2) + z - 4 = 0$ $z^2 + z - 6 = 0$

This is a quadratic equation in $z$. Factoring it, we get: $(z+3)(z-2) = 0$ The possible values for $z$ are $z = -3$ or $z = 2$.

Now we relate $z$ back to $y$. We must remember that $y > 0$. For $y > 0$, by the AM-GM inequality, $y + \frac{1}{y} \ge 2\sqrt{y \cdot \frac{1}{y}} = 2$. So, $z = y + \frac{1}{y}$ must satisfy $z \ge 2$.

Case 1: $z = -3$ $y + \frac{1}{y} = -3$. This value $z=-3$ is less than 2. As expected, this case yields no positive solutions for $y$. Multiplying by $y$ gives $y^2 + 1 = -3y$, or $y^2 + 3y + 1 = 0$. The discriminant is $\Delta = 3^2 - 4(1)(1) = 9 - 4 = 5 > 0$. The roots are $y = \frac{-3 \pm \sqrt{5}}{2}$. Since $\sqrt{5} \approx 2.236$, both roots $\frac{-3 + \sqrt{5}}{2} \approx \frac{-0.764}{2}$ and $\frac{-3 - \sqrt{5}}{2} \approx \frac{-5.236}{2}$ are negative. Since $y$ must be positive, $z=-3$ yields no valid solutions for $y$.

Case 2: $z = 2$ $y + \frac{1}{y} = 2$. This value $z=2$ satisfies $z \ge 2$. Multiplying by $y$ gives $y^2 + 1 = 2y$, or $y^2 - 2y + 1 = 0$. This equation factors as $(y-1)^2 = 0$. The only solution is $y = 1$. This is a positive value for $y$.

So, the only positive root of the polynomial in $y$ is $y=1$. Now we convert back to $x$ using $y = e^x$: $e^x = 1$ $x = \ln(1)$ $x = 0$

This is a real root for the original equation. Since $y=1$ was the only positive root for the polynomial in $y$, there is only one real root for the original equation.

To confirm, let's check $x=0$ in the original equation: $e^{4(0)} + e^{3(0)} - 4e^{2(0)} + e^0 + 1 = e^0 + e^0 - 4e^0 + e^0 + 1 = 1 + 1 - 4(1) + 1 + 1 = 1 + 1 - 4 + 1 + 1 = 4 - 4 = 0$. So $x=0$ is indeed a root.

The number of real roots is 1.