The number of real roots of the equation e4x $-$ e3x $-$ 4e2x $-$ ex + 1 = 0 is equal to ______________.
Answer (integer)
2
Solution
t<sup>4</sup> $-$ t<sup>3</sup> $-$ 4t<sup>2</sup> $-$ t + 1 = 0, e<sup>x</sup> = t > 0<br><br>$\Rightarrow {t^2} - t - 4 - {1 \over t} + {1 \over {{t^2}}} = 0$<br><br>$\Rightarrow {\alpha ^2} - \alpha - 6 = 0,\alpha = t + {1 \over t} \ge 2$<br><br>$\Rightarrow \alpha = 3, - 2$ (reject)<br><br>$\Rightarrow t + {1 \over t} = 3$<br><br>$\Rightarrow$ The number of real roots = 2
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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