Question

Solve (5+2root6) ^x2-3 + (5+2root6) ^x2-3 =10 by using quadratic formula. How is this solved?

Answer 1

$$\begin{gathered} \mathrm{Assuming}\mathrm{your}\mathrm{question}\mathrm{as}, \\ {\left(5+2\sqrt{6}\right)}^{{\mathrm{x}}^2-3}+{\left(5+2\sqrt{6}\right)}^{{\mathrm{x}}^2-3}=10 \\ 2{\left(5+2\sqrt{6}\right)}^{{\mathrm{x}}^2-3}=10 \\ {\left(5+2\sqrt{6}\right)}^{{\mathrm{x}}^2-3}=5 \\ \mathrm{Taking}\log \mathrm{we}\mathrm{get}, \\ \log {\left(5+2\sqrt{6}\right)}^{{\mathrm{x}}^2-3}=\log 5 \\ \left({\mathrm{x}}^2-3\right)\log \left(5+2\sqrt{6}\right)=\log 5 \\ {\mathrm{x}}^2-3=\frac{\log 5}{\log \left(5+2\sqrt{6}\right)}={\log }_{\left(5+2\sqrt{6}\right)}5 \\ {\mathrm{x}}^2={\log }_{\left(5+2\sqrt{6}\right)}5+3 \\ \mathrm{x}=\sqrt{\left(\left({\log }_{\left(5+2\sqrt{6}\right)}5\right)+3\right)} \end{gathered}$$