Question

Resolvendo a equação modular |2x+6| + |- x+3| = 9, teremos como solução valores de x, cuja soma será igual a:
a:3
b:4
c:6
d:-2
e:5

Answer 1

$\large\boxed{\begin{array}{l}\underline{\rm D~\!\!efinic_{\!\!,}\tilde ao\,de\,m\acute odulo}\\\sf |x|=\begin{cases}\sf~~ x,~~se~~x\geqslant0\\\sf -x,~~se~x<0\end{cases}\\\\\sf |2x+6|+|-x+3|=9\\\rm Vamos\,analisar\,a\,soma\,dos\,m\acute odulos\\\rm em\,tr\hat es\,intervalos:\\\rm\bullet 1\,no\,intervalo\, x<-3\\\rm\bullet 2\,no\,intervalo -3\leqslant x\leqslant3\\\rm\bullet 3\,no\,intervalo \,x>3\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm soma\,dos\,m\acute odulos\,no\,intervalo\,x<-3:}\\\sf A\,express\tilde ao |2x+6|\,torna-se\,-2x-6\,e\,a\,express\tilde ao\\\sf |-x+3|\,torna-se\,-x+3\\\sf DA\acute I\\\sf -2x-6-x+3=9\\\sf -3x-3=9\\\sf 3x=-3-9\\\sf 3x=-12\\\sf x=-\dfrac{12}{3}\\\\\sf x=-4\\\sf como~-4<0\,ent\tilde ao\, S_1=\{-4\}\end{array}}$  

$\large\boxed{\begin{array}{l}\underline{\rm Soma\,dos\,m\acute odulos\,no\,intervalo -3\leqslant x\leqslant3:}\\\sf A\,express\tilde ao\,|2x+6|\,torna-se\,2x+6\,e\,a\,express\tilde ao\\\sf |-x+3|\,torna-se\, -x+3\\\sf DA\acute I\\\sf 2x+6-x+3=9\\\sf x+9=9\\\sf x=9-9\\\sf x=0\\\sf como\,0<3\,ent\tilde ao~S_2=\{0\}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Soma\,dos\,m\acute odulos\,no\,intervalo\,x>3:}\\\sf A\,express\tilde ao\, |2x+6|\,torna-se\,2x+6\\\sf e\,a\,express\tilde ao\,|-x+3|\,torna-se\,x-3.\\\sf DA\acute I\\\sf 2x+6+(x-3)=9\\\sf 2x+6+x-3=9\\\sf 3x+3=9\\\sf 3x=9-3\\\sf 3x=6\\\sf x=\dfrac{6}{3}\\\\\sf x=2\\\sf como~2<3~ent\tilde ao\, S_3=\bigg\{\bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\sf Como\,o\,exerc\acute icio\,pede\,a\,soma\,das\,ra\acute izes\\\sf ent\tilde ao\\\sf-4+0=-4 \end{array}}$