Răspuns:
[tex]x \notin \mathbb{R}[/tex]
Explicație pas cu pas:
[tex]2\sqrt{3} \,x^2+9\sqrt{3} +\sqrt{12} \,x^2=0[/tex]
[tex]2\sqrt{3} \,x^2+9\sqrt{3} +2\sqrt{3} \,x^2=0[/tex]
[tex]2\sqrt{3} \, x^2+2\sqrt{3} x^2+9\sqrt{3} =0[/tex]
[tex]4\sqrt{3} x^2+9\sqrt{3} =0\,|:\sqrt{3}[/tex]
[tex]\implies 4x^2+9=0\implies 4x^2=-9[/tex]
[tex]\implies x^2=-\frac{9}{4} \implies x \notin \mathbb{R}[/tex]
