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Resolvendo as equações primeiro:
( 1 )
[tex]\begin{array}{l}\sf 12x^2=0\\\\\sf \dfrac{12x^2}{12}=\dfrac{0}{12}\\\\\sf x^2=0\\\\\sf \sqrt{x^2}=\sqrt{0}\\\\\sf |x|=0\\\\\sf \therefore~x'=x''~\Rightarrow~0\end{array}[/tex]
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( 2 )
[tex]\begin{array}{l}\sf x^2-64=0\\\\\sf 64+x^2-64=0+64\\\\\sf x^2=64\\\\\sf \sqrt{x^2}=\sqrt{64}\\\\\sf |x|=8\\\\\sf x=\pm~8\\\\\sf \therefore~x'=8\quad e\quad x''=-8\end{array}[/tex]
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( 3 )
[tex]\begin{array}{l}\sf x^2-1=0\\\\\sf 1+x^2-1=0+1\\\\\sf x^2=1\\\\\sf \sqrt{x^2}=\sqrt{1}\\\\\sf |x|=1\\\\\sf x=\pm~1\\\\\sf \therefore~x'=1\quad e\quad x''=-1\end{array}[/tex]
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( 4 )
[tex]\begin{array}{l}\sf x^2-2x=0\\\\\sf x(x-2)=0\\\\\begin{cases}\sf x=0\\\\\sf x-2=0\end{cases}\\\\\begin{cases}\sf x'=0\\\\\sf x''=2\end{cases}\end{array}[/tex]
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( 5 )
[tex]\begin{array}{l}\sf 12x^2-48=0\\\\\sf 48+12x^2-48=0+48\\\\\sf 12x^2=48\\\\\sf \dfrac{12x^2}{12}=\dfrac{48}{12}\\\\\sf x^2=4\\\\\sf \sqrt{x^2}=\sqrt{4}\\\\\sf |x|=2\\\\\sf x=\pm~2\\\\\sf \therefore~x'=2\quad e\quad x''=-2\end{array}[/tex]
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Assim relacionando a primeira coluna com a segunda:
( 2 ) => S = {8 ; -8}
( 1 ) => S = { 0 }
( 4 ) => S = {0 ; 2}
( 3 ) => S = {1 ; - 1}
( 5 ) => S = {2 ; -2}
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RESPOSTA: dessa forma a alternativa que se encaixa é: Letra D)
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