Question

If $(2+\sqrt{3}{)}^{x^2-2x+1}+(2-\sqrt{3}{)}^{x^2-2x-1}=\frac{2}{2-\sqrt{3}}$, then x is equal to:

• A. 1
• B. 2
• C. both (a) and (c)

Answer 1

The correct option is C both (a) and (c)

$\because 2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$
$\therefore$ the given equation can be expressed as
$(2+\sqrt{3}{)}^{x^2-2x+1}+\frac{1}{(2+\sqrt{3}{)}^{x^2-2x+1}}=2(2+\sqrt{3})$ . . .(i)
Let us put $y=(2+\sqrt{3}{)}^{x^2-2x+1}$, then the eq.(i) can be expressed as
$y+\frac{(2+\sqrt{3}{)}^2}{y}=2(2+\sqrt{3})\Rightarrow y^2-2(2+\sqrt{3})y+(2+\sqrt{3}{)}^2=0\Rightarrow [y-(2+\sqrt{3}){]}^2=0\Rightarrow y=2+\sqrt{3}\therefore (2+\sqrt{3}{)}^{x^2-2x+1}=(2+\sqrt{3})\Rightarrow x^2-2x+1=1\Rightarrow x(x-2)=0\Rightarrow x=0or2$