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

The given equation is $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$. Let $y = e^x$. Since $x$ is a real number, $y = e^x > 0$. Substituting $y = e^x$ 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 $y = e^x$ must be positive. For each positive real root $y_0$ of this equation, there is a unique real root $x_0 = \ln(y_0)$ for the original equation.

The equation $y^4 + y^3 - 4y^2 + y + 1 = 0$ is a reciprocal equation because the coefficients are symmetric (1, 1, -4, 1, 1). Since $y=e^x > 0$, $y \neq 0$, we can divide the equation by $y^2$:

$\frac{y^4}{y^2} + \frac{y^3}{y^2} - \frac{4y^2}{y^2} + \frac{y}{y^2} + \frac{1}{y^2} = 0$

$y^2 + y - 4 + \frac{1}{y} + \frac{1}{y^2} = 0$

Group the 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 grouped equation:

$(z^2 - 2) + z - 4 = 0$

$z^2 + z - 6 = 0$

This is a quadratic equation in $z$. Factoring the quadratic equation:

$(z+3)(z-2) = 0$

The possible values for $z$ are $z = -3$ or $z = 2$.

Now we relate $z$ back to $y$. Recall $z = y + \frac{1}{y}$. Since $y = e^x > 0$, we can use the AM-GM inequality: For $y > 0$, $y + \frac{1}{y} \ge 2\sqrt{y \cdot \frac{1}{y}} = 2$. Equality holds when $y = \frac{1}{y}$, which means $y^2 = 1$. Since $y>0$, $y=1$. Thus, for $y>0$, $z = y + \frac{1}{y}$ must be greater than or equal to 2.

Case 1: $z = -3$ This value $z=-3$ is less than 2. This means there are no positive real values of $y$ that satisfy $y + \frac{1}{y} = -3$. (Solving $y^2 + 3y + 1 = 0$ gives $y = \frac{-3 \pm \sqrt{5}}{2}$, both of which are negative).

Case 2: $z = 2$ This value $z=2$ is equal to 2, which is possible for $y>0$. $y + \frac{1}{y} = 2$ Multiply by $y$: $y^2 + 1 = 2y$ $y^2 - 2y + 1 = 0$ $(y-1)^2 = 0$

This gives a single root $y = 1$. This root $y=1$ is positive, so it is a valid value for $y = e^x$.

Now we find the corresponding value(s) of $x$:

$y = e^x = 1$

Taking the natural logarithm of both sides:

$x = \ln(1)$

$x = 0$

This is a real root. We have found only one positive real root for $y$ (which is $y=1$), which corresponds to exactly one real root for $x$ (which is $x=0$).

To confirm that $y=1$ is the only positive root of $y^4 + y^3 - 4y^2 + y + 1 = 0$, we can see that for $y>0$, the expression $g(y) = (y+3)(y-2)$ where $z=y+1/y$. Since $z \ge 2$ for $y>0$, $z+3 > 0$. If $z=2$, $g(y)=(2+3)(2-2)=0$, which occurs when $y=1$. If $z>2$ (i.e., $y>0, y \neq 1$), then $z-2 > 0$, so $g(y) = (z+3)(z-2) > 0$. Thus, $y=1$ is the only positive root of $y^4 + y^3 - 4y^2 + y + 1 = 0$.

Since $y=e^x$ must be positive, the only solution for $y$ that comes from $y^4 + y^3 - 4y^2 + y + 1 = 0$ and is positive is $y=1$.

$e^x = 1$ gives $x=0$.

Therefore, the original equation has exactly one real root.