Question

A) |x|^2 =5|x|-6
B) 6.(2^2x^-15)>48

Preciso urgente!!!!
Por favor pelo menos a letra a

Answer 1

A) $|x|^2=5|x|-6$


Faça a seguinte mudança de variável:

$|x|=t~~~~~(t\ge 0)$


Substituindo, a equação fica

$t^2=5t-6\\\\ t^2-5t+6=0~~~\Rightarrow~~\left\{ \!\begin{array}{l} a=1\\b=-5\\c=6 \end{array} \right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=(-5)^2-4\cdot 1\cdot 6\\\\ \Delta=25-24\\\\ \Delta=1$


$t=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ t=\dfrac{-(-5)\pm \sqrt{1}}{2\cdot 1}\\\\\\ t=\dfrac{5\pm 1}{2}\\\\\\ \begin{array}{rcl} t=\dfrac{5-1}{2}&~\text{ ou }~&t=\dfrac{5+1}{2}\\\\ t=\dfrac{4}{2}&~\text{ ou }~&t=\dfrac{6}{2}\\\\ t=2&~\text{ ou }~&t=3 \end{array}$


Subsituindo de volta para a variável $x,$ ficamos com

$\begin{array}{rcl} |x|=2&~\text{ ou }~&|x|=3 \end{array}\\\\\ \boxed{\begin{array}{c} \begin{array}{rcl} x=\pm\,2&~\text{ ou }~&x=\pm\,3 \end{array} \end{array}}$


Conjunto solução: $S=\{-3,\,-2,\,2,\,3\}.$

______________

B) $6\cdot (2^{2x}-15)>48$

$2^{2x}-15>\dfrac{48}{6}\\\\\\ 2^{2x}-15>8\\\\ 2^{2x}>8+15\\\\ 2^{2x}>23\\\\ 2x>\mathrm{\ell og}_2\,23\\\\ \boxed{\begin{array}{c}x>\dfrac{1}{2}\,\mathrm{\ell og}_2\,23 \end{array}}$


Bons estudos! :-)