Question

If ${\left( {5 + 2\sqrt 6 } \right)^{\left( {{x^2} - 3} \right)}} + {\left( {5 - 2\sqrt 6 } \right)^{\left( {{x^2} - 3} \right)}} = 10$ , the $x$ is equal to
(A). $ \pm 3$ or \[ \pm \sqrt 3 \]
(B). $ \pm 5$ or \[\sqrt 5 \]
(C). $ \pm 4$ or \[\sqrt 4 \]
(D). $ \pm 2$ or \[ \pm \sqrt 2 \]

Answer 1

Hint: In the solution we will use rationalization and substitution methods. The concept of the rationalization method is to remove the radical symbols or any number in the denominator.

Complete Step-by-step Solution
Given: ${\left( {5 + 2\sqrt 6 } \right)^{\left( {{x^2} - 3} \right)}} + {\left( {5 - 2\sqrt 6 } \right)^{\left( {{x^2} - 3} \right)}} = 10$
We will rationalizing $\left( {5 - 2\sqrt 6 } \right)$ by dividing and multiplying by $\left( {5 - 2\sqrt 6 } \right)$, we get the value gives as,
$\begin{array}{c}
{\left[ {\left( {5 - 2\sqrt 6 } \right)\dfrac{{\left( {5 + 2\sqrt 6 } \right)}}{{\left( {5 + 2\sqrt 6 } \right)}}} \right]^{\left( {{x^2} - 3} \right)}} = \dfrac{{25 - 24}}{{\left( {5 + 2\sqrt 6 } \right)}}\\
 = {\left( {\dfrac{1}{{5 + 2\sqrt 6 }}} \right)^{\left( {{x^2} - 3} \right)}}
\end{array}$
Now, on putting the value of $\left( {5 - 2\sqrt 6 } \right)$ in the above expression, we will get,
${\left( {5 + 2\sqrt 6 } \right)^{\left( {{x^2} - 3} \right)}} + {\left( {\dfrac{1}{{\left( {5 + 2\sqrt 6 } \right)}}} \right)^{\left( {{x^2} - 3} \right)}} = 10$
Let us assume that \[{\left( {5 + 2\sqrt 6 } \right)^{\left( {{x^2} - 3} \right)}} = t\], then we have,
$\begin{array}{c}
t + \dfrac{1}{t} = 10\\
{t^2} - 10t + 1 = 0
\end{array}$
On solving the quadratic equation ${t^2} - 10t + 1 = 0$, we will get the value as,
$\begin{array}{c}
t = \dfrac{{10 \pm \sqrt {100 - 4} }}{2}\\
t = \dfrac{{10 \pm \sqrt {96} }}{2}\\
t = \dfrac{{10 \pm 4\sqrt 6 }}{2}\\
t = 5 \pm 2\sqrt 6
\end{array}$
Therefore, we can say that the roots of the quadratic equation ${t^2} - 10t + 1 = 0$ are $5 + 2\sqrt 6 $ and $5 - 2\sqrt 6 $.
On simplifying the roots of the equation, we get,
${\left( {5 + 2\sqrt 6 } \right)^{\left( {{x^2} - 3} \right)}} = 5 + 2\sqrt 6 $
On comparing the powers of both the sides of the above expression, we get,
$\begin{array}{c}
{x^2} - 3 = 1\\
{x^2} = 4\\
x = \pm 2
\end{array}$
Also, again simplifying the roots of the equation, we get,
$\begin{array}{c}
{\left( {5 + 2\sqrt 6 } \right)^{\left( {{x^2} - 3} \right)}} = 5 - 2\sqrt 6 \\
{\left( {5 + 2\sqrt 6 } \right)^{\left( {{x^2} - 3} \right)}} = {\left( {5 + 2\sqrt 6 } \right)^{ - 1}}
\end{array}$
On comparing the powers of both the sides of the above expression, we get,
$\begin{array}{l}
{x^2} - 3 = - 1\\
{x^2} = - 4\\
x = \pm \sqrt 2
\end{array}$
Therefore, the correct option is (D) that is $ \pm 2$ or \[ \pm \sqrt 2 \].

Note: Make sure to use a rationalization method when you will see this type of questions. The tricky part is comparing the powers of the equations.