[tex]\it 7+4\sqrt3=4+3+4\sqrt3=2^2+4\sqrt3+(\sqrt3)^2=(2+\sqrt3)^2\\ \\ Analog, 7-4\sqrt3=(2-\sqrt3)^2\\ \\ Dar,\ \ 2-\sqrt3=\dfrac{1}{2+\sqrt3}[/tex]
Acum, ecuația devine:
[tex]\it(2+\sqrt3)^{2x} +\dfrac{1}{(2+\sqrt3)^{2x}} =4\\ \\ \\ Notez\ \ (2+\sqrt3)^{2x}=t,\ cu\ t>0,\ iar\ ecua\c{\it t}ia\ devine:\\ \\ t+\dfrac{1}{t}=4 \Rightarrow t^2-4t+1=0\Rightarrow t^2-4t+4-3=0\Rightarrow (t-2)^2-(\sqrt3)^2=0\Rightarrow\\ \\ \Rightarrow(t-2-\sqrt3)(t-2+\sqrt3)=0\Rightarrow t_1=2-\sqrt3,\ \ t_2=2+\sqrt3[/tex]
Revenim asupra notației:
[tex]\it I)\ \ t=2-\sqrt3=\dfrac{1}{2+\sqrt3}=(2+\sqrt3)^{-1} \Rightarrow(2+\sqrt3)^{2x}=(2+\sqrt3)^{-1}\Rightarrow\\ \\ \\ \Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\\ \\ \\ II)\ \ t=2+\sqrt3=(2+\sqrt3)^1 \Rightarrow (2+\sqrt3)^{2x}=(2+\sqrt3)^1\Rightarrow2x=1\Rightarrow\\ \\ \Rightarrow x=\dfrac{1}{2}[/tex]
Prin urmare, ecuația dată admite două soluții:
[tex]\it x_1=-\dfrac{1}{2},\ \ x_2=\dfrac{1}{2}[/tex]
