Question

The solution set of the equation ${\left(5+2\sqrt{6}\right)}^{x^2-3}+{\left(5-2\sqrt{6}\right)}^{x^2-3}=10$ is

• A. $\left\{\pm 2,\pm \sqrt{2}\right\}$
• B. $\left\{\pm 3,\pm \sqrt{3}\right\}$
• C. $\left\{\pm 5,\pm \sqrt{5}\right\}$
• D. $\left\{\pm 6,\pm \sqrt{6}\right\}$

Answer 1

The correct option is A

$\left\{\pm 2,\pm \sqrt{2}\right\}$

The explanation for the correct option:

Step-1 Modify the equation :

The given equation: ${\left(5+2\sqrt{6}\right)}^{x^2-3}+{\left(5-2\sqrt{6}\right)}^{x^2-3}=10$

Let us assume that, ${\left(5+2\sqrt{6}\right)}^{x^2-3}=m$.

Now, ${\left(5-2\sqrt{6}\right)}^{x^2-3}=\frac{{\left(5-2\sqrt{6}\right)}^{x^2-3}{\left(5+2\sqrt{6}\right)}^{x^2-3}}{{\left(5+2\sqrt{6}\right)}^{x^2-3}}$

$\begin{array}{rcl}& \Rightarrow & {\left(5-2\sqrt{6}\right)}^{x^2-3}=\frac{{\left[\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)\right]}^{x^2-3}}{{\left(5+2\sqrt{6}\right)}^{x^2-3}}\\ & \Rightarrow & {\left(5-2\sqrt{6}\right)}^{x^2-3}=\frac{{\left[{\left(5\right)}^2-{\left(2\sqrt{6}\right)}^2\right]}^{x^2-3}}{{\left(5+2\sqrt{6}\right)}^{x^2-3}}\\ & \Rightarrow & {\left(5-2\sqrt{6}\right)}^{x^2-3}=\frac{{\left[25-24\right]}^{x^2-3}}{{\left(5+2\sqrt{6}\right)}^{x^2-3}}\\ & \Rightarrow & {\left(5-2\sqrt{6}\right)}^{x^2-3}=\frac{{\left[1\right]}^{x^2-3}}{m}\left[\because {\left(5+2\sqrt{6}\right)}^{x^2-3}=m\right]\\ & \Rightarrow & {\left(5-2\sqrt{6}\right)}^{x^2-3}=\frac{1}{m}\end{array}$

Step-2 Solution of equation:

Thus, consider the equation: ${\left(5+2\sqrt{6}\right)}^{x^2-3}+{\left(5-2\sqrt{6}\right)}^{x^2-3}=10$

$$\begin{gathered} \Rightarrow m+\frac{1}{m}=10 \\ \Rightarrow \frac{m^2+1}{m}=10 \\ \Rightarrow m^2-10m+1=0 \\ \Rightarrow m=\frac{-\left(-10\right)\pm \sqrt{{\left(-10\right)}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)} \\ \Rightarrow m=\frac{10\pm \sqrt{100-4}}{2} \\ \Rightarrow m=\frac{10\pm \sqrt{96}}{2} \\ \Rightarrow m=\frac{10\pm 4\sqrt{6}}{2} \\ \Rightarrow m=5\pm 2\sqrt{6} \end{gathered}$$

Now, $m={\left(5+2\sqrt{6}\right)}^{x^2-3}=\frac{1}{{\left(5-2\sqrt{6}\right)}^{x^2-3}}={\left(5-2\sqrt{6}\right)}^{-x^2+3}$

Thus, ${\left(5+2\sqrt{6}\right)}^{x^2-3}=\left(5+2\sqrt{6}\right)$

$$\begin{gathered} \Rightarrow x^2-3=1 \\ \Rightarrow x^2=4 \\ \Rightarrow x=\pm 2 \end{gathered}$$

Thus, ${\left(5-2\sqrt{6}\right)}^{-x^2+3}=\left(5-2\sqrt{6}\right)$

$$\begin{gathered} \Rightarrow -x^2+3=1 \\ \Rightarrow x^2=2 \\ \Rightarrow x=\pm \sqrt{2} \end{gathered}$$

Therefore, the solution set of the equation ${\left(5+2\sqrt{6}\right)}^{x^2-3}+{\left(5-2\sqrt{6}\right)}^{x^2-3}=10$ is $\left\{\pm 2,\pm \sqrt{2}\right\}$.

Hence, option A is correct.