[tex]\it 4+2\sqrt3 = 3+1+2\sqrt3=(\sqrt3)^2+2\sqrt3+1^2 =(\sqrt3+1)^2\ \ \ \ (*)
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4-2\sqrt3 = 3+1-2\sqrt3=(\sqrt3)^2-2\sqrt3+1^2 =(\sqrt3-1)^2[/tex]
[tex]\it \sqrt{(\sqrt3+1)^2} = |\sqrt3+1| = \sqrt3+1, deoarece\ \sqrt3+1\ \textgreater \ 0
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\sqrt{(\sqrt3-1)^2} = |\sqrt3-1| = \sqrt3-1, deoarece\ \sqrt3-1\ \textgreater \ 0[/tex]
Expresia din enunț devine:
[tex]\it (\sqrt3+1+\sqrt3-1)^2+4= (2\sqrt3)^2 +4 = 12+4 = 16 = 4^2[/tex]
