The sum of all the roots of the equation e^2x-11e^x-45e^-x+812=0 is _e p.

Question

The sum of all the roots of the equation $e^{2x}-11e^x-45e^{-x}+\dfrac{81}2=0$ is $\log_e p$. Then $p$ is equal to \_\_\_\_

Answer: 45

Accepted Answer

Let $t=e^x$. Multiplying through by $2t$:
$2t^3-22t^2+81t-90=0.$

If the roots (in $t$) are $e^{x_1},e^{x_2},e^{x_3}$, then by the product-of-roots relation for this cubic:
$$e^{x_1}\cdot e^{x_2}\cdot e^{x_3}=\frac{90}2=45 \implies e^{x_1+x_2+x_3}=45 \implies x_1+x_2+x_3=\ln45.$$

So $p=45$.

Solutions · 1