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Solve for $x$, $\left(5+2\sqrt{6}\right)^{x^2-3}+\left(5-2\sqrt{6}\right)^{x^2-3}=10$

Last updated date: 24th Jul 2024
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Hint: We are given an equation in irrationals and we need to simplify it and find the value of the unknown variable $x$. We need to apply some trick involving the irrationals here. We will use the concept of inverses in irrationals and we will consider one variable as $y$ and then the other term will become $\dfrac{1}{y}$. After doing this we will get an equation in a different variable $y$ and then we will solve for that, then re-substitute it in the older variable to get the value.

Complete step by step answer:
Let $y=\left(5+2\sqrt{6}\right)^{x^2-3}$
Then, if we multiply the quantity inside bracket by 1 i.e. $\dfrac{5-2\sqrt{6}}{5-2\sqrt{6}}$, then we get:
$\left(5+2\sqrt{6}\right)\times \dfrac{5-2\sqrt{6}}{5-2\sqrt{6}}=\dfrac{25-24}{5-2\sqrt{6}}=\dfrac{1}{5-2\sqrt{6}}$
$\implies \dfrac{1}{y}=5-2\sqrt{6}$
So, the equation becomes:
$\implies y^2+1=10y$
$\implies y^2-10y+1=0$
Solving this equation using the discriminant method:
For any general quadratic equation:
$ay^2+by+c=0$, the roots of this equation will be given by:
$y=\dfrac{10\pm \sqrt{100-4}}{2\times 1}=\dfrac{10\pm \sqrt{96}}{2}$
$\implies y=\dfrac{10\pm 4\sqrt{6}}{2}$
$\implies y=5\pm 2\sqrt{6}$
Putting back the value of $y$ in the equation above:
Or we have:
$\implies \left(5+2\sqrt{6}\right)^1=\left(5+2\sqrt{6}\right)^{x^2-3}$, or
${{\left( 5-2\sqrt{6} \right)}^{-1}}={{\left( 5+2\sqrt{6} \right)}^{{{x}^{2}}-3}}$
$\implies x^2-3=1$ or $x^2-3=-1$
$\implies x^2=4$ or $x^2=2$
$\implies x=\pm 2$ or $x=\pm\sqrt{2}$
Hence, the value of $x$ has been found out.

Note: Do not forget the $\pm$ sign while writing for the solution of any quadratic equation because then you might miss some of the solutions. You should remember the fact that the quadratic equation has two solutions if the discriminant is non-zero, which is the case here. Moreover, while dealing with the irrationals make sure you make no calculation mistakes.