How many real values?

Let ${ (5+2\sqrt { 6 } ) }^{ x^{ 2 }-3 }$ =t

then equation becomes ${ t }^{ 2 }-10t+1=0$

So $t={ (5\pm 2\sqrt { 6 } ) }$

Hence ${ x }^{ 2 }-3=\pm 1$ . So 4 values of x.

$(5+2\sqrt{6})^{x^{2}-3} + (5-2\sqrt{6})^{x^{2}-3} = 10$

Notice that $5+2\sqrt{6}$ and $5-2\sqrt{6}$ look like solutions to a quadratic equation. Let us us say;

$p = 5 \pm 2\sqrt{6}$

So the equation in the question resolves to:

$p^{x^2 -3} + p^{x^2 -3} = 10$

$p^{x^2 -3} = 5$

$\log_{p}{5} = x^2 - 3$

$x= \pm\sqrt{\log_{p}{5}+3}$

$x= \pm\sqrt{\log_{5 \pm 2\sqrt{6}}{5}+3}$

Now, notice that in this equation there are 2 plus/minus signs, which are mutually inexclusive, meaning that there are four possible combinations of the ordered plus/minuses. Thus there exist four solutions. We know that both the solutions for $p$ are real, and so a quick check that the inequality $\log_{p}{5}+3 > 0$ holds true will reveal that all four solutions are real.

Please note that $5-2\sqrt{6}=\frac{1}{5+2\sqrt{6}}$ . So the equation above can be rewritten as

$(5+2\sqrt{6})^{x^2-3} +\frac{1}{(5+2\sqrt{6})^{x^2-3}}=10$ .

Considering a new variable $a=(5+2\sqrt{6})^{x^2-3}$ , we find $a+\frac{1}{a}=10$ , or $a^2-10a+1=0$ , whose solutions are $a_1=5+2\sqrt{6}$ and $a_2=5-2\sqrt{6}=(5+2\sqrt{6})^{-1}$ .

Returning to the original variable $x$ , we get either

$(5+2\sqrt{6})^{x^2-3}=(5+2\sqrt{6})^1$ , that is, $x^2-3=1$ , whose solutions are $x=\pm 2$

or

$(5+2\sqrt{6})^{x^2-3}=(5+2\sqrt{6})^{-1}$ , that is, $x^2-3=-1$ , whose solutions are $x=\pm\sqrt{2}$ .

So the solution set is $S=\{\pm\sqrt{2},\pm 2\}$ , which has 4 elements.

Just observe the question carefully and you would get the solution

Just observe that second is conjugate of first. Assume the first one to be t. then solve for t. You will get two values of t and each will correspond to 2 values of x !!!