Question

The total number of integral values of $x$, which satisfy ${(\sqrt{3}+1)}^{2x}+{(\sqrt{3}-1)}^{2x}=2^{3x}$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$(\sqrt{3}+1{)}^{2x}+(\sqrt{3}-1{)}^{2x}=2^{3x}$
$\Rightarrow (\sqrt{3}+1{)}^{2x}\left[1+{\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)}^{2x}\right]=2^{3x}$
$\Rightarrow \left[1+{\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)}^{2x}\right]=\frac{2^{3x}}{{(\sqrt{3}+1)}^{2x}}$
$\Rightarrow 1+{\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\right)}^{2x}=\frac{2^{3x}}{{(\sqrt{3}+1)}^{2x}}$ ........... rationalization

$\Rightarrow 1+{\left(\frac{3+1-2\sqrt{3}}{2}\right)}^{2x}=\frac{8^x}{{(3+1+2\sqrt{3})}^x}$ ............. Using $(a-b{)}^2=a^2-2ab+b^2$ and $a^2-b^2=(a-b)(a+b)$
$\Rightarrow 1+(2-\sqrt{3}{)}^{2x}={\left(\frac{4}{2+\sqrt{3}}\right)}^x$
$\Rightarrow 1+(2-\sqrt{3}{)}^{2x}={\left(\frac{4}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}\right)}^x$ ........... rationalization
$\Rightarrow 1+(2-\sqrt{3}{)}^{2x}={\left(\frac{8-4\sqrt{3}}{4-3}\right)}^x$ ............. Using $a^2-b^2=(a-b)(a+b)$
$\Rightarrow 1+(4+3-4\sqrt{3}{)}^x=(8-4\sqrt{3}{)}^x$

$\Rightarrow 1+(7-4\sqrt{3}{)}^x=(8-4\sqrt{3}{)}^x\left[so,thevalueofxalways1\right]$
$\therefore x=1$
$so,thecorrectoptionisA.$