Mathematics - Quadratic Equation Question with Solution | TestHub

Given:

$S=\left\{x:{\left(\sqrt{3}+\sqrt{2}\right)}^{x^2-4}+{\left(\sqrt{3}-\sqrt{2}\right)}^{x^2-4}=10\right\}$

Now,

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

So,

${\left(\sqrt{3}+\sqrt{2}\right)}^{x^2-4}+{\left(\sqrt{3}-\sqrt{2}\right)}^{x^2-4}=10$

$\Rightarrow {\left(\sqrt{3}+\sqrt{2}\right)}^{x^2-4}+{\left(\frac{1}{\sqrt{3}+\sqrt{2}}\right)}^{x^2-4}=10$

Put ${\left(\sqrt{3}+\sqrt{2}\right)}^{x^2-4}=u$, then

$u+\frac{1}{u}=10$

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

$\Rightarrow u=\frac{10\pm \sqrt{100-4}}{2}$

$\Rightarrow u=5\pm \sqrt{24}$

$\Rightarrow u=5\pm 2\sqrt{3}\sqrt{2}$

$\Rightarrow u={\left(\sqrt{3}\pm \sqrt{2}\right)}^2$

$\Rightarrow {\left(\sqrt{3}+\sqrt{2}\right)}^{x^2-4}={\left(\sqrt{3}\pm \sqrt{2}\right)}^2$

So, ${\left(\sqrt{3}+\sqrt{2}\right)}^{x^2-4}={\left(\sqrt{3}+\sqrt{2}\right)}^2$ and ${\left(\sqrt{3}+\sqrt{2}\right)}^{x^2-4}={\left(\frac{1}{\sqrt{3}+\sqrt{2}}\right)}^2$

Therefore,

$x^2-4=2 \& x^2-4=-2$

$\Rightarrow x=\pm \sqrt{6} \& x=\pm \sqrt{2}$

So, $S=\left\{\sqrt{6}, \sqrt{2}, -\sqrt{6}, -\sqrt{2}\right\}$

Hence, $n\left(S\right)=4$