Question

Solutions of $(2+\sqrt{3}{)}^{x^2-2x+1}+(2-\sqrt{3}{)}^{x^2-2x-1}=\frac{4}{2-\sqrt{3}}$ are

• A. $1\pm \sqrt{3},1$
• B. $1\pm \sqrt{2},1$
• C. $1\pm \sqrt{3},2$
• D. $1\pm \sqrt{2},2$

Answer 1

The correct option is B $1\pm \sqrt{2},1$

$(2+\sqrt{3}{)}^{x^2-2x+1}+(2-\sqrt{3}{)}^{x^2-2x-1}=\frac{4}{2-\sqrt{3}}$
$\Rightarrow (2+\sqrt{3}{)}^{x^2-2x}+(2-\sqrt{3}{)}^{x^2-2x}=4$
Let $(2+\sqrt{3}{)}^{x^2-2x}=t$
$\Rightarrow t+\frac{1}{t}=4$
$\Rightarrow t^2-4t+1=0$
$t=2\pm \sqrt{3}$
$\Rightarrow x^2-2x=1$ or $x^2-2x=-1$
$\therefore x=1\pm \sqrt{2}$ or $x=1$
Hence, option B .