Question

Resolva a equação:
$\sqrt{x^2-4x+4} +|x + 4|=3$
Postei somente para quem gosta de esquentar a cuca.

Answer 1

Resposta:

√(x² - 4x + 4) + lx + 4| = 3

lx - 2l + lx + 4l = 3

não existe solução

Answer 2

$\Large\boxed{\begin{array}{l}\underline{\rm D~\!\!efinic_{\!\!,}\tilde oes~de~m\acute odulo}\\\\\boxed{\tt 1}~\sf |x|=\sqrt{x^2}\\\\\boxed{\tt 2}~\sf|x|=\begin{cases}\sf ~~x,~se~x\geqslant0\\\sf-x,~se~x<0\end{cases}\\\underline{\rm Fatorac_{\!\!,}\tilde ao~do~trin\hat omio}\\\underline{\rm quadrado~perfeito}\\\sf a^2\pm2ab+b^2=(a\pm b)^2\end{array}}$

$\Large\boxed{\begin{array}{l}\sf\sqrt{x^2-4x+4}+|x+4|=3\\\sf vamos~escrever~o~radical~\sqrt{x^2-4x+4}\\\sf usando~a~d~\!\!efinic_{\!\!,}\tilde ao~de~m\acute odulo.\\\sf x^2-4x+4=(x-2)^2\\\sf \sqrt{x^2-4x+4}=\sqrt{(x-2)^2}=|x-2|\\\sf voltando~a~express\tilde ao~teremos:\\\sf |x-2|+|x+4|=3\\\sf vamos~analisar~a~soma~dos~m\acute odulos\\\sf em~tr\hat es~intervalos:\\\sf\cdot 1~no~intervalo~x\leqslant-4\\\sf\cdot 2~no~intervalo~-4\leqslant x<2\\\sf\cdot 3~no~intervalo~x\geqslant2\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\boldsymbol{Algumas~considerac_{\!\!,}\tilde oes}}\\\underline{\boldsymbol{~sobre~os~m\acute odulos~envolvidos}}\\\underline{\boldsymbol{na~quest\tilde ao.}}\\\sf |x-2|=\begin{cases}\sf x-2,~se~x\geqslant2\\\sf -(x-2)=2-x,~se~x<2\end{cases}\\\\\sf|x+4|=\begin{cases}\sf x+4,~se~x\geqslant-4\\\sf-(x+4)=-x-4,~se~x<-4\end{cases}\end{array}}$$\large\boxed{\begin{array}{l}\underline{\rm soma~dos~m\acute odulos~no~intervalo~x\leqslant-4:}\\\sf a~express\tilde ao~|x-2|~torna~se~2-x\\\sf e~a~express\tilde ao~|x+4|~torna-se~-x-4.\\\sf ou~seja,\\\sf2-x+(-x-4)=3\\\sf 2-x-x-4=3\\\sf -2x-2=3\\\sf -2x=3+2\\\sf -2x=5\cdot(-1)\\\sf 2x=-5\\\sf x=-\dfrac{5}{2}\\\sf n\tilde ao~temos~soluc_{\!\!,}\tilde ao~neste~intervalo\\\sf pois~-\dfrac{5}{2}>-4\\\sf s_1=\bigg\{\bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Soma~dos~m\acute odulos~no~intervalo~-4\leqslant x<2:}\\\sf a~express\tilde ao~|x-2|~torna-se~2-x\\\sf e~a~express\tilde ao~|x+4|~torna-se~x+4\\\sf isto~\acute e,\\\sf 2-x+x+4=3\\\sf 0x=3-4-2\\\sf 0x=-3\\\sf s_2=\bigg\{\bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Soma~dos~m\acute odulos~no~intervalo~x\geqslant2:}\\\sf a~express\tilde ao~|x-2|~torna-se~x-2\\\sf e~a~express\tilde ao~|x+4|~torna-se~x+4\\\sf ent\tilde ao,\\\sf x-2+x+4=3\\\sf 2x=3+2-4\\\sf 2x=1\\\sf x=\dfrac{1}{2}\\\sf n\tilde ao~temos~soluc_{\!\!,}\tilde ao~neste~intervalo\\\sf pois~\dfrac{1}{2}<2\\\sf s_3=\bigg\{\bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm A~soluc_{\!\!,}\tilde ao~geral~\acute e~obtida}\\\underline{\rm fazendo~as~intersecc_{\!\!,}\tilde oes}\\\underline{\rm dos~conjuntos~soluc_{\!\!,}\tilde oes.}\\\sf S= s_1\cap s_2\cap s_3\\\sf como~todos~s\tilde ao~vazios,resta~afirmar~que\\\sf S=\bigg\{\bigg\}\end{array}}$