Question

The number of real solutions of the equation (5 + 2$\sqrt{6})$${}^{x^2-3}$ + (5 - 2$\sqrt{6})$${}^{x^2-3}$ = 10 is

• A. 2
• B. 4
• C. 6
• D. none of these

Answer 1

The correct option is B 4

Given equation (5 + 2$\sqrt{6})$${}^{x^2-3}$ + (5 - 2$\sqrt{6})$${}^{x^2-3}$ = 10 $\dots \left(1\right)$

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

$\Rightarrow \frac{{(5+2\sqrt{6})}^{x^2-3}{(5-2\sqrt{6})}^{x^2-3}}{{(5-2\sqrt{6})}^{x^2-3}}=t$
$\Rightarrow \frac{{\left[(5+2\sqrt{6})(5-2\sqrt{6})\right]}^{x^2-3}}{{(5-2\sqrt{6})}^{x^2-3}}=t$

$\Rightarrow \frac{{[25-24]}^{x^2-3}}{{(5-2\sqrt{6})}^{x^2-3}}=t$

$\Rightarrow \frac{1}{{(5-2\sqrt{6})}^{x^2-3}}=t$

$\Rightarrow {(5-2\sqrt{6})}^{x^2-3}=\frac{1}{t}$

$from\left(1\right)$
$\therefore t+\frac{1}{t}=10$

$\Rightarrow t^2-10t+1=0$

$\Rightarrow t=\frac{10\pm \sqrt{100.4}}{2}$
$\Rightarrow t=\frac{10\pm 4\sqrt{6}}{2}=5\pm 2\sqrt{6}$
$\therefore {(5+2\sqrt{6})}^{x^2-3}=5+2\sqrt{6}$

$\Rightarrow x^2-3=1$

$\Rightarrow x^2=4\Rightarrow x=\pm 2$

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

$\Rightarrow x^2-3=-1\Rightarrow x^2=2$

$\Rightarrow x=\pm \sqrt{2}.$

Hence, number of solution are $4$