Question

O produto das raízes da equação |3x + 5| + |x - 1| = 2

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}\end{array}}$$\boxed{\begin{array}{l}\sf Vamos~analisar~a~soma~dos~m\acute odulos~em~3~intervalos:\\\sf\cdot 1~no~intervalo~x\leqslant-\dfrac{5}{3}\\\sf\cdot 2~no~intervalo~-\dfrac{5}{3}\leqslant x<1\\\sf\cdot 3~no~intervalo~x\geqslant1\end{array}}$

$\boxed{\begin{array}{l}\sf No~intervalo~x\leqslant-\dfrac{5}{3}\\\sf a~express\tilde ao~|3x+5|~se~torna~negativa~e~a~express\tilde ao\\\sf |x-1|~tamb\acute em\\\sf pela~d~\!\!efinic_{\!\!,}\tilde ao~de~m\acute odulo~podemos~escrever:\\\sf-(3x+5)-(x-1)=2\\\sf -3x-5-x+1=2\\\sf -4x-4=2\\\sf -4x=2+4\\\sf -4x=6\cdot(-1)\\\sf 4x=-6\\\sf x=-\dfrac{6\div2}{4\div2}=-\dfrac{3}{2}\\\sf por\acute em~-\dfrac{3}{2}~n\tilde ao~pertence~ao~intervalo~porque\\\sf-\dfrac{3}{2}>-\dfrac{5}{3}\end{array}}$

$\boxed{\begin{array}{l}\sf Portanto~no~1^o~intervalo~n\tilde ao~h\acute a~soluc_{\!\!,}\tilde ao.\end{array}}$

$\boxed{\begin{array}{l}\sf No~intervalo~-\dfrac{5}{3}\leqslant x<1\\\sf a~express\tilde ao~|3x+5|~\acute e~positiva~e~a~express\tilde ao\\\sf |x-1|~\acute e~negativa.\\\sf pela~d~\!\!efinic_{\!\!,}\tilde ao~de~m\acute odulo~podemos~escrever:\\\sf 3x+5-(x-1)=2\\\sf 3x+5-x+1=2\\\sf 2x+6=2\\\sf 2x=2-6\\\sf 2x=-4\\\sf x=-\dfrac{4}{2}\\\sf x=-2\\\sf -2~n\tilde ao~pertence~ao~intervalo~dado~pois~\\\sf -\dfrac{5}{3}>-2\end{array}}$

$\boxed{\begin{array}{l}\sf Portanto~no~2^o~intervalo~n\tilde ao~h\acute a~soluc_{\!\!,}\tilde oes.\end{array}}$

$\boxed{\begin{array}{l}\sf No~intervalo~x\geqslant1\\\sf a~express\tilde ao~|3x+5|~\acute e~positiva\\\sf e~a~express\tilde ao~|x-1|~tamb\acute em.\\\sf pela~d~\!\!efinic_{\!\!,}\tilde ao~de~m\acute odulo~podemos~escrever:\\\sf3x+5+x-1=2\\\sf 4x+4=2\\\sf4x=2-4\\\sf 4x=-2\\\sf x=-\dfrac{2}{4}\\\sf x=-\dfrac{1}{2}\\\sf por\acute em~-\dfrac{1}{2}~n\tilde ao~pertence~ao~intervalo~dado\\\sf o ~que~nos~leva~a~concluir~que~n\tilde ao~h\acute a~soluc_{\!\!,}\tilde ao\\\sf no~mesmo\end{array}}$

$\large\boxed{\begin{array}{l}\rm Conclus\tilde ao:\\\sf a~equac_{\!\!,}\tilde ao~|3x+5|+|x-1|=2\\\sf n\tilde ao~admite~soluc_{\!\!,}\tilde ao~portanto~\\\sf S=\bigg\{\bigg\}\end{array}}$