Question

Qual o produto das raízes de equação exponencial anexada?

Answer 1

$3^x+\dfrac{1}{3^x}=\dfrac{4\sqrt{3}}{3}$


Vamos fazer uma mudança de variável:

$3^x=t~~~~(t>0)$


Substituindo, a equação fica

$t+\dfrac{1}{t}=\dfrac{4\sqrt{3}}{3}\\\\\\ \dfrac{t^2}{t}+\dfrac{1}{t}=\dfrac{4\sqrt{3}}{3}\\\\\\ \dfrac{t^2+1}{t}=\dfrac{4\sqrt{3}}{3}\\\\\\ 3\cdot (t^2+1)=4\sqrt{3}\,t\\\\ 3t^2+3=4\sqrt{3}\,t$

$3t^2-4\sqrt{3}\,t+3=0~~~\Rightarrow~~\left\{\!\begin{array}{l}a=3\\b=-4\sqrt{3}\\c=3 \end{array}\right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=\big(4\sqrt{3}\big)^2-4\cdot 3\cdot 3\\\\ \Delta=4^2\cdot 3-36\\\\ \Delta=48-36\\\\ \Delta=12=\\\\ \Delta=2^2\cdot 3$

$t=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ t=\dfrac{-\big(-4\sqrt{3}\big)\pm \sqrt{2^2\cdot 3}}{2\cdot 3}\\\\\\ t=\dfrac{4\sqrt{3}\pm 2\sqrt{3}}{2\cdot 3}\\\\\\ t=\dfrac{\diagup\!\!\!\! 2\sqrt{3}\cdot (2\pm 1)}{\diagup\!\!\!\! 2\cdot 3}\\\\\\ t=\dfrac{\sqrt{3}\cdot (2\pm 1)}{3}$

$t=\dfrac{\sqrt{3}\cdot (2\pm 1)}{3}\\\\\\ \begin{array}{rcl} t=\dfrac{\sqrt{3}\cdot (2-1)}{3}&~\text{ ou }~&t=\dfrac{\sqrt{3}\cdot (2+1)}{3}\\\\ t=\dfrac{\sqrt{3}\cdot 1}{3}&~\text{ ou }~&t=\dfrac{\sqrt{3}\cdot \diagup\!\!\!\! 3}{\diagup\!\!\!\! 3}\\\\ t=\dfrac{\sqrt{3}}{3}&~\text{ ou }~&t=\sqrt{3} \end{array}$


Voltando à variável original $x,$ temos

$\begin{array}{rcl} 3^x=\dfrac{\sqrt{3}}{3}&~\text{ ou }~&3^x=\sqrt{3}\\\\\\ 3^x=\dfrac{3^{1/2}}{3}&~\text{ ou }~&3^x=3^{1/2}\\\\\\ 3^x=3^{1/2}\cdot 3^{-1}&~\text{ ou }~&3^x=3^{1/2}\\\\\\ 3^x=3^{(1/2)-1}&~\text{ ou }~&3^x=3^{1/2}\\\\\\ 3^x=3^{-1/2}&~\text{ ou }~&3^x=3^{1/2} \end{array}$


Portanto,

$\begin{array}{rcl} x=-\dfrac{1}{2}&~\text{ ou }~&x=\dfrac{1}{2} \end{array}$


As raízes da equação são

$x_1=-\dfrac{1}{2}~\text{ e }~x_2=\dfrac{1}{2}$


de forma que o produto entre elas é

$x_1\cdot x_2\\\\ =-\dfrac{1}{2}\cdot \dfrac{1}{2}\\\\\\ =\boxed{\begin{array}{c}-\dfrac{1}{4} \end{array}}$


Bons estudos! :-)