Question

What is the number of real roots of the equation \[{e^{4x}} + {e^{3x}} - 4{e^{2x}} + {e^x} + 1 = 0\]?
A. \[3\]
B. \[4\]
C. \[1\]
D. \[2\]

Answer 1

Hint: Divide both sides of the given equation by \[{e^{2x}}\]. You’ll get a quadratic equation in \[\left( {{e^x} + \dfrac{1}{{{e^x}}}} \right)\]. Make the substitution \[{e^x} + \dfrac{1}{{{e^x}}} = y\] and find the value of \[y\]. You’ll get two values of \[y\]. Omit the negative value because \[{e^x} > 0\] for all values of \[x\]. Replace \[y\] by \[\left( {{e^x} + \dfrac{1}{{{e^x}}}} \right)\]. You’ll get a quadratic equation in \[{e^x}\]. Find the roots by solving the equation. After that the value of \[x\] will be obtained.

Formula Userd:
\[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
\[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]

Complete step-by-step solution:
The given equation is \[{e^{4x}} + {e^{3x}} - 4{e^{2x}} + {e^x} + 1 = 0\]
Dividing each term by \[{e^{2x}}\] on both sides and using the formula \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\], we get
\[{e^{2x}} + {e^x} - 4 + \dfrac{1}{{{e^x}}} + \dfrac{1}{{{e^{2x}}}} = 0\]
Arrange the terms.
\[ \Rightarrow \left( {{e^{2x}} + \dfrac{1}{{{e^{2x}}}}} \right) + \left( {{e^x} + \dfrac{1}{{{e^x}}}} \right) - 4 = 0\]
Use the formula \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[ \Rightarrow \left\{ {{{\left( {{e^x}} \right)}^2} + {{\left( {\dfrac{1}{{{e^x}}}} \right)}^2}} \right\} + \left( {{e^x} + \dfrac{1}{{{e^x}}}} \right) - 4 = 0\]
Use the identity \[{a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab\]
\[ \Rightarrow \left\{ {{{\left( {{e^x} + \dfrac{1}{{{e^x}}}} \right)}^2} - 2 \times {e^x} \times \dfrac{1}{{{e^x}}}} \right\} + \left( {{e^x} + \dfrac{1}{{{e^x}}}} \right) - 4 = 0\]
\[ \Rightarrow {\left( {{e^x} + \dfrac{1}{{{e^x}}}} \right)^2} - 2 + \left( {{e^x} + \dfrac{1}{{{e^x}}}} \right) - 4 = 0\]
\[ \Rightarrow {\left( {{e^x} + \dfrac{1}{{{e^x}}}} \right)^2} + \left( {{e^x} + \dfrac{1}{{{e^x}}}} \right) - 6 = 0\]
Substitute \[{e^x} + \dfrac{1}{{{e^x}}} = y\]
Then the equation reduces to \[{y^2} + y - 6 = 0\]
Solve this equation by factorizing the expression of the left-hand side.
\[ \Rightarrow {y^2} + 3y - 2y - 6 = 0\]
\[ \Rightarrow y\left( {y + 3} \right) - 2\left( {y + 3} \right) = 0\]
\[ \Rightarrow \left( {y + 3} \right)\left( {y - 2} \right) = 0\]
Product of two factors is equal to zero if and only if any of the two factors is equal to zero.
So, either \[y + 3 = 0\] or \[y - 2 = 0\]
\[y + 3 = 0 \Rightarrow y = - 3\]
\[y - 2 = 0 \Rightarrow y = 2\]
But \[y = - 3\] is not possible because \[{e^x} > 0\] for all values of \[x\]
So, \[y = 2\]
We assumed \[{e^x} + \dfrac{1}{{{e^x}}} = y\]
So, \[{e^x} + \dfrac{1}{{{e^x}}} = 2\]
Multiply each term by \[{e^x}\] on both sides of the equation.
\[ \Rightarrow {\left( {{e^x}} \right)^2} + 1 = 2{e^x}\]
Let us substitute \[{e^x} = z\]
Then the equation reduces to \[{z^2} + 1 = 2z\]
This is a quadratic equation in \[z\].
Solve this equation to find the value of \[z\]
Arrange the equation.
\[{z^2} - 2z + 1 = 0\]
Use the identity \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
\[ \Rightarrow {\left( {z - 1} \right)^2} = 0\]
\[ \Rightarrow z - 1 = 0\]
\[ \Rightarrow z = 1\]
We assumed that \[{e^x} = z\]
So, \[{e^x} = 1\]
Any non-zero number raise to the power zero is equal to \[1\].
So, we can write \[{e^0} = 1\]
\[\therefore {e^x} = {e^0}\]
\[ \Rightarrow x = 0\]
So, only one value of \[x\] is obtained.
Thus, number of real roots of the given equation is \[1\].
Hence option C is correct.

Note: Remember that the value of any exponential function is positive always. Many times a critical equation can be solved by making a proper substitution. So, be careful while making a substitution.