The number of real roots of the equation,
e^4x + e^3x – 4e^2x + e^x + 1 = 0 is :

e^4x + e^3x – 4e^2x + e^x + 1 = 0 <br><br>Dividing by e^2x, we get <br><br>e^2x + e^x - 4 + ${1 \over {{e^x}}}$ + ${1 \over {{e^{2x}}}}$ = 0 <br><br>$\Rightarrow$ $$\left( {{e^{2x}} + {1 \over {{e^{2x}}}}} \right) + \left( {{e^x} + {1 \over {{e^x}}}} \right)$$ - 4 = 0 <br><br>$\Rightarrow$ ${\left( {{e^x} + {1 \over {{e^x}}}} \right)^2} - 2$ + $\left( {{e^x} + {1 \over {{e^x}}}} \right)$ - 4 = 0 <br><br>Let ${{e^x} + {1 \over {{e^x}}} = z}$ <br><br>(z² – 2) + (z) – 4 = 0 <br><br>$\Rightarrow$ z² + z – 6 = 0 <br><br>$\Rightarrow$ z = –3, 2 <br><br>$\therefore$ ${{e^x} + {1 \over {{e^x}}} = 2}$ <br><br>$\Rightarrow$ (e^x – 1)² = 0 $\Rightarrow$ x = 0. <br><br>$\therefore$ Number of real roots = 1