Question

The number of real roots of the equation $(\sqrt{3}+1{)}^{2x}+(\sqrt{3}-1{)}^{2x}=2^{3x}$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. more than $2$

Answer 1

The correct option is B $1$

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

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

Now, $\sqrt{3}+1<2\sqrt{2}$

$\Rightarrow \frac{\sqrt{3}+1}{2\sqrt{2}}<1$

Substitute $\frac{\sqrt{3}+1}{2\sqrt{2}}=\sin \theta$

$\Rightarrow \cos \theta =\sqrt{1-{\sin }^2\theta }\Rightarrow \cos \theta =\sqrt{1-\frac{4+2\sqrt{3}}{8}}=\sqrt{\frac{4-2\sqrt{3}}{8}}=\sqrt{{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}^2}=\frac{\sqrt{3}-1}{2\sqrt{2}}$

Equation $\left(1\right)$ becomes $(\sin \theta {)}^{2x}+(\cos \theta {)}^{2x}=1$

$x=1$ is a real solution of the above equation .