Question

$(2+ \sqrt{3} )^x+((2- \sqrt{3} )^x=4$

Answer 1

$2- \sqrt{3} = \frac{(2- \sqrt{3})(2+ \sqrt{3}) }{2+ \sqrt{3} } =\frac{1}{2+ \sqrt{3} }$

$(2+ \sqrt{3})^x+ \frac{1}{(2+ \sqrt{3})^x } =4$
$(2+ \sqrt{3} )^x=t$, t>0
$t+ \frac{1}{t} -4=0$
t²-4t+1=0
D=2√3
t₁=2+√3
t₂=2-√3

$(2+ \sqrt{3} )^x=2+ \sqrt{3}$
x=1
$(2+ \sqrt{3} )^x=2- \sqrt{3}$
x=-1

Answer 2

$2- \sqrt{3} = \frac{(2- \sqrt{3})(3+ \sqrt{3} ) )}{2+ \sqrt{3} } = \frac{1}{2+ \sqrt{3} }$

$(2+ \sqrt{3} ) ^{x} + \frac{1}{(2+ \sqrt{3} ) ^{x}} =4$
$(2+ \sqrt{3} )^x=b$, $b \ \textgreater \ 0$

$b^2-4b+1=0$
$D=2 \sqrt{3}$   
D - дискриминант
$b _{1} =2+ \sqrt{3}$
$b _{2} =2- \sqrt{3}$
$(2+ \sqrt{3} )^x=2+ \sqrt{3}$
$x=1$
$(2+ \sqrt{3} )^x=2- \sqrt{3}$
$x=-1$