Verified Answer :

Lilith

\(f (x) = x^2 - 3 \\ \\ a.\ 6 \\ \\f(x)=6\\ \\ 6=x^2-3\\ \\6+3= x^2\)

\(x^2=9 \\ \\x^2-9=0\\ \\(x-3)(x+3)=0 \\ \\x-3=0 \ \ or \ \ x+3 =0 \\ \\x=3 \ \ or \ \ x= -3 \\ \\ Answer : \ Possible .\)


\(b.\ \ 0 \\ \\f(x)=0\\ \\ 0=x^2-3\\ \\(x-\sqrt{3}) (x+\sqrt{3})=0 \\ \\ x-\sqrt{3}=0 \ \ or \ \x+\sqrt{3} =0\)

\(x=\sqrt{3} \ \ or \ \x=-\sqrt{3} \\ \\Answer : \ x \ is \ not \ an \ integer,\ not \ possible\)


\(c.\ \ -6 \\ \\f(x)= -6\\ \\ -6=x^2-3\\ \\ x^2=-6 +3 \\ \\x^2=-3\\ \\Answer: \ square \ of \ an \ integer \ can \ not \ be \ negative\)



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