Question

solution of equation ( 5+2√6)^{​x^2-3} + ( 5-2√6 )^{x^2-3} = 10 are

Answer 1

Dear student,

We can see that we could solve this equation by cancelling out $2\sqrt{6}$ from the equation.
We can solve this question by inspection. Let us plug in x = $\pm$2 and see what happens.

$$\begin{gathered} (5+\hspace{0.17em}2\sqrt{6}{)}^{{\mathrm{x}}^2-3}+(5-2\sqrt{6}{)}^{{\mathrm{x}}^2-3}=10 \\ \mathrm{Putting}\mathrm{x}=2,\mathrm{we}\mathrm{get}: \\ (5+2\sqrt{6}{)}^1+\hspace{0.17em}(5-2\sqrt{6}{)}^1=10 \\ 10=10,\mathrm{which}\mathrm{is}\mathrm{true}. \\ \mathrm{Similarly}\mathrm{if}\mathrm{we}\mathrm{put}\mathrm{x}=-2,\mathrm{we}\mathrm{get} \\ (5+2\sqrt{6}{)}^1+\hspace{0.17em}(5-2\sqrt{6}{)}^1=10 \\ 10=10,\mathrm{which}\mathrm{is}\mathrm{true} \\ \mathrm{Apart}\mathrm{of}\mathrm{these}\mathrm{values}\mathrm{of}\mathrm{x},\mathrm{the}\mathrm{equation}\mathrm{is}\mathrm{not}\mathrm{true}\mathrm{for}\mathrm{any}\mathrm{other}\mathrm{integer}. \\ \mathrm{Hence},\mathrm{integer}\mathrm{solution}\mathrm{for}\mathrm{this}\mathrm{equation}\mathrm{will}\mathrm{be}:\hspace{0.17em} \\ \mathrm{x}=\pm 2 \end{gathered}$$

Regards