Official solution
e4x + e3x -- 4e2x + ex + 1 = 0
Dividing by e2x, we get
e2x + ex - 4 + \[{1 \over {{e^x}}}\] + \[{1 \over {{e^{2x}}}}\] = 0
\[ \Rightarrow \] \[\left( {{e^{2x}} + {1 \over {{e^{2x}}}}} \right) + \left( {{e^x} + {1 \over {{e^x}}}} \right)\] - 4 = 0
\[ \Rightarrow \] \[{\left( {{e^x} + {1 \over {{e^x}}}} \right)^2} - 2\] + \[\left( {{e^x} + {1 \over {{e^x}}}} \right)\] - 4 = 0
Let \[{{e^x} + {1 \over {{e^x}}} = z}\]
(z2
-- 2) + (z) -- 4 = 0
\[ \Rightarrow \] z2
+ z -- 6 = 0
\[ \Rightarrow \] z = --3, 2
\[ \therefore \] \[{{e^x} + {1 \over {{e^x}}} = 2}\]
\[ \Rightarrow \] (ex -- 1)2 = 0 \[ \Rightarrow \] x = 0.
\[ \therefore \] Number of real roots = 1